The graph of the piecewise-linear map oscillating between $0$ and $1$ on the intervals $[1\mathrm{/}(n+2),1\mathrm{/}(n+1)]$ is path-connected, its closure adds the segment $\{0\}\times [0,1]$, and that closure is connected, is not path-connected because no path joins the segment to the graph, and is not locally connected

$f$ is continuous. For $n\in \mathbb{N}$ the restriction of $f$ to the closed set $I_n\cup I_{n+1}=[1\mathrm{/}(n+3),1\mathrm{/}(n+1)]$ is continuous, being affine on each of the two closed pieces by [A1] and agreeing at the shared endpoint, so the finite closed cover clause of [A2] applies with two pieces; likewise $f$ is affine, hence continuous, on $[1\mathrm{/}2,1]=I_0$.

$(0,1]$ is order-convex, hence a connected subset of $\mathbb{R}$ by [A4], and it is path-connected: for $x,y\in (0,1]$ the map $t\mapsto x+t(y-x)$ is continuous by [A1] and takes values in $(0,1]$ by order-convexity.

$(0,1]$ is locally connected: a basic open subset of it is the trace of an interval of $\mathbb{R}$, hence order-convex, hence connected by [A4]; so the open connected subsets form a neighbourhood base at each of its points, which is [A9].

$\bar{G}\subseteq G\cup (\{0\}\times [0,1])$. Let $(x,y)\in \bar{G}$. Every point of $G$ lies in $[0,1]\times [0,1]$, which is closed, being a product of closed sets whose complement is a union of basic open sets; so $(x,y)\in [0,1]\times [0,1]$ by [A5]. If $x=0$ the point lies in $\{0\}\times [0,1]$.

A second consequence, used twice below: for every real $\rho >0$ there are $x_0,x_1\in (0,\rho ]$ with $f(x_0)=0$ and $f(x_1)=1$. Indeed [A7] gives a natural $k\ge 1$ with $1\mathrm{/}k<\rho$; with $n:=k-1\in \mathbb{N}$ the interval $I_n\subseteq (0,\rho ]$, and the two endpoints of $I_n$ carry the values ${\epsilon }_n$ and ${\epsilon }_{n+1}$, which are $0$ and $1$ in one order or the other.

The sets $W_n:=(1\mathrm{/}(n+3),1\mathrm{/}(n+1))$ for $n\in \mathbb{N}$, together with $W^{\ast }:=(1\mathrm{/}2,1]$, form an open cover of $(0,1]$ in its subspace topology, and $f$ restricted to each is a restriction of one of the continuous maps of step 1.1; so $f$ is continuous by the open cover clause of [A2].

The map $g:(0,1]\to {\mathbb{R}}^2$, $g(x):=(x,f(x))$, is continuous by [A3], its components being the inclusion and $f$, both continuous by step 2.1 and [A2]; it is injective, since $g(x)$ determines $x$; and its image is $G$.

$\{0\}\times [0,1]\subseteq \bar{G}$. Let $t\in [0,1]$ and let $(a,b)\times (c,d)$ be a basic open set containing $(0,t)$. By [A7] there is a natural $k\ge 1$ with $1\mathrm{/}k<\min \{b,1\}$, and putting $n:=k-1\in \mathbb{N}$ the interval $I_n$ lies in $(0,b)$, since $1\mathrm{/}(n+1)=1\mathrm{/}k<b$. As $f$ runs affinely between $0$ and $1$ on $I_n$, [A4] gives $x\in I_n$ with $f(x)\in (c,d)$: the image of $I_n$ under $f$ is order-convex and contains $0$ and $1$, hence contains $t$ and every point near it inside $[0,1]$. Then $(x,f(x))\in G$ lies in the basic set, so $(0,t)\in \bar{G}$ by [A5].

The corestriction $g_0:(0,1]\to G$ is a continuous bijection by step 3.1 and [A2], and its inverse is the restriction of ${\pi }_0$ to $G$, which is continuous by [A3] and [A2]; so $g_0$ is a homeomorphism and $G\cong (0,1]$.

Suppose instead $x>0$, so $x\in (0,1]$ and $f(x)$ is defined. Let $\epsilon >0$. By step 2.1 and [A3] there is $\delta >0$ such that $\mathrm{\mid }f(s)-f(x)\mathrm{\mid }<\epsilon$ for every $s\in (x-\delta ,x+\delta )\cap (0,1]$. The basic set $(x-\delta ,x+\delta )\times (y-\epsilon ,y+\epsilon )$ contains $(x,y)$, hence meets $G$ by [A5] in a point $(s,f(s))$; then $\mathrm{\mid }y-f(x)\mathrm{\mid }\le \mathrm{\mid }y-f(s)\mathrm{\mid }+\mathrm{\mid }f(s)-f(x)\mathrm{\mid }<2\epsilon$. As $\epsilon >0$ was arbitrary, $y=f(x)$ and $(x,y)\in G$. With step 3.2 and step 1.4 this proves claim 2.

Claim 5. Fix $t\in [0,1]$ and let $U:=\bar{G}\cap (\mathbb{R}\times (t-1\mathrm{/}4,t+1\mathrm{/}4))$, an open subset of $\bar{G}$ containing $(0,t)$. Suppose $V$ is open in $\bar{G}$, connected, with $(0,t)\in V\subseteq U$. Then $V$ contains a set $\bar{G}\cap ((-\eta ,\eta )\times (t-\eta ,t+\eta ))$ for some $\eta >0$ by [A3], and that set meets $G$ by step 3.2, at a point whose first coordinate $x$ satisfies $0<x<\eta$.

Hence $G$ is path-connected, connected and locally connected, these being carried across the homeomorphism of step 4.1 from step 1.2 and step 1.3, using [A6] for connectedness and [A9] for local connectedness. This is claim 1.

A useful consequence of claim 2, used twice below: if $p\in \bar{G}$ has ${\pi }_0(p)>0$ then $p=({\pi }_0(p),f({\pi }_0(p)))$, so ${\pi }_1(p)=f({\pi }_0(p))$.

Claim 4. Suppose $\gamma :[0,1]\to \bar{G}$ is a path with $\gamma (0)\in \{0\}\times [0,1]$ and $\gamma (1)\in G$, and write $k:={\pi }_0\circ \gamma$ and $h:={\pi }_1\circ \gamma$, both continuous by [A3] and [A2]. Then $J:=\{u\in [0,1]:k(u)=0\}$ is closed in $[0,1]$, being the preimage of the closed set $\{0\}$, it contains $0$, and $1\mathrm{\notin }J$ since ${\pi }_0(\gamma (1))>0$.

Claim 3: $G$ is connected by step 5.1 and $G\subseteq \bar{G}\subseteq \bar{G}$, so $\bar{G}$ is connected by [A6].

Let $c:=\sup J$, which exists by [A8]. Every open set containing $c$ contains an interval around it, which by [A8] meets $J$; so $c\in \bar{J}$, and $J$ is closed in $[0,1]$ while $\bar{J}\subseteq \overline{[0,1]}=[0,1]$, so $c\in J$ by [A5]. Hence $k(c)=0$ and $c<1$.

So ${\pi }_0[V]$ is a connected subset of $\mathbb{R}$ by [A2, A3, A4], hence order-convex, and it contains $0$ and $x>0$; therefore $[0,x]\subseteq {\pi }_0[V]$. By step 1.5 with $\rho :=x$ there are $x_0,x_1\in (0,x]$ with $f(x_0)=0$ and $f(x_1)=1$, so $V$ contains points $p_0,p_1$ with ${\pi }_0(p_i)=x_i>0$, and ${\pi }_1(p_i)=f(x_i)$ by step 5.2.

By continuity of $h$ at $c$ there is $\delta >0$ with $c+\delta \le 1$ and $\mathrm{\mid }h(u)-h(c)\mathrm{\mid }<1\mathrm{/}4$ for all $u\in [c,c+\delta ]$; put $t:=h(c)$. Moreover $k(c+\delta )>0$, since $c+\delta >c=\sup J$ puts $c+\delta$ outside $J$ while $k\ge 0$ everywhere by step 1.4.

The restriction of $k$ to $[c,c+\delta ]$ is continuous on a connected space by [A4] and step 1.2, so its image is order-convex and contains $k(c)=0$ and $k(c+\delta )>0$; hence $[0,k(c+\delta )]$ lies in that image. By step 1.5 with $\rho :=k(c+\delta )$ there are $x_0,x_1\in (0,k(c+\delta )]$ with $f(x_0)=0$ and $f(x_1)=1$, and therefore $u_0,u_1\in [c,c+\delta ]$ with $k(u_i)=x_i>0$.

By step 5.2, $h(u_i)=f(k(u_i))=f(x_i)$, so $h(u_0)=0$ and $h(u_1)=1$; but both lie within $1\mathrm{/}4$ of $t$ by step 7.1, giving $1=\mathrm{\mid }h(u_1)-h(u_0)\mathrm{\mid }\le \mathrm{\mid }h(u_1)-t\mathrm{\mid }+\mathrm{\mid }t-h(u_0)\mathrm{\mid }<1\mathrm{/}2$, which is false. So no such path exists, and since $\bar{G}$ contains points of both kinds by claim 2, it is not path-connected. This is claim 4.

Hence $V$ contains a point with second coordinate $0$ and a point with second coordinate $1$, both of which must lie in $(t-1\mathrm{/}4,t+1\mathrm{/}4)$ because $V\subseteq U$; that gives $1\le \mathrm{\mid }0-t\mathrm{\mid }+\mathrm{\mid }t-1\mathrm{\mid }<1\mathrm{/}2$, which is false. So no such $V$ exists and $\bar{G}$ is not locally connected at $(0,t)$, by [A9]; this is claim 5.