The graph of the piecewise-linear map oscillating between $0$ and $1$ on the intervals $[1/(n+2), 1/(n+1)]$ 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 $I_n \cup I_{n+1} = [1/(n+3), 1/(n+1)]$ 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 $[1/2,1] = I_0$.

$(0,1]$ 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].

$\overline{G} \subseteq G \cup (\{0\} \times [0,1])$. Let $(x,y) \in \overline{G}$. 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 $1/k < \rho$; with $n := k-1 \in \mathbb{N}$ the interval $I_n \subseteq (0, \rho]$, and the two endpoints of $I_n$ carry the values $\varepsilon_n$ and $\varepsilon_{n+1}$, which are $0$ and $1$ in one order or the other.

The sets $W_n := (1/(n+3), 1/(n+1))$ for $n \in \mathbb{N}$, together with $W^{*} := (1/2, 1]$, 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 $G$.

$\{0\} \times [0,1] \subseteq \overline{G}$. 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 $1/k < \min\{b, 1\}$, and putting $n := k - 1 \in \mathbb{N}$ the interval $I_n$ lies in $(0, b)$, since $1/(n+1) = 1/k < b$. 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 $(0,t) \in \overline{G}$ 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 $\varepsilon > 0$. By step 2.1 and [A3] there is $\delta > 0$ such that $|f(s) - f(x)| < \varepsilon$ for every $s \in (x-\delta, x+\delta) \cap (0,1]$. The basic set $(x-\delta, x+\delta) \times (y-\varepsilon, y+\varepsilon)$ contains $(x,y)$, hence meets $G$ by [A5] in a point $(s, f(s))$; then $|y - f(x)| \le |y - f(s)| + |f(s) - f(x)| < 2\varepsilon$. As $\varepsilon > 0$ 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 $U := \overline{G} \cap (\mathbb{R} \times (t - 1/4,\ t + 1/4))$, an open subset of $\overline{G}$ containing $(0,t)$. Suppose $V$ is open in $\overline{G}$, connected, with $(0,t) \in V \subseteq U$. Then $V$ contains a set $\overline{G} \cap ((-\eta,\eta) \times (t-\eta, t+\eta))$ 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 $p \in \overline{G}$ 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 $\gamma : [0,1] \to \overline{G}$ 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 $1 \notin J$ since $\pi_0(\gamma(1)) > 0$.

Claim 3: $G$ is connected by step 5.1 and $G \subseteq \overline{G} \subseteq \overline{G}$, so $\overline{G}$ 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 $c \in \overline{J}$, and $J$ is closed in $[0,1]$ while $\overline{J} \subseteq \overline{[0,1]} = [0,1]$, 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 $|h(u) - h(c)| < 1/4$ 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 $1/4$ of $t$ by step 7.1, giving $1 = |h(u_1) - h(u_0)| \le |h(u_1) - t| + |t - h(u_0)| < 1/2$, which is false. So no such path exists, and since $\overline{G}$ 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 $(t - 1/4, t+1/4)$ because $V \subseteq U$; that gives $1 \le |0 - t| + |t - 1| < 1/2$, which is false. So no such $V$ exists and $\overline{G}$ is not locally connected at $(0,t)$, by [A9]; this is claim 5.