The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies placed in the connectedness hierarchy

Discrete. If $X$ has two distinct points $x,y$ then $(\{x\},X\setminus \{x\})$ is a separation by [A2] and [A1], so $X$ is disconnected; every connected subset therefore has at most one point, the subspace topology on a subset again being discrete, so the components are the singletons by [A3], and $X$ is not path-connected, a path being in particular a connected image. Every singleton is open and connected, so $X$ is locally connected by [A3].

Indiscrete. The only open sets are $\mathrm{\varnothing }$ and $X$ by [A2], so no two nonempty open sets are disjoint and $X$ is connected by [A1]; hence, for nonempty $X$, its only component is $X$ by [A3], and $X$ itself is an open connected set containing every point, so $X$ is locally connected by [A3].

Indiscrete, path-connectedness. Every function $\gamma :[0,1]\to X$ is continuous by [A4], the preimages of $\mathrm{\varnothing }$ and $X$ being $\mathrm{\varnothing }$ and $[0,1]$; so for $x,y\in X$ the function with $\gamma (0)=x$ and $\gamma (t)=y$ for $t>0$ is a path from $x$ to $y$, and $X$ is path-connected.

Cofinite, $X$ infinite. Let $U,V$ be nonempty open sets, with complements $C_U$, $C_V$ finite by [A2]. Then $X\setminus (U\cap V)=C_U\cup C_V$ is finite by [A2], so $U\cap V\mathrm{\ne }\mathrm{\varnothing }$, $X$ being infinite. Hence no separation exists and $X$ is connected by [A1].

Cocountable, $X$ uncountable, and $X$ a subset of $\mathbb{R}$ or any uncountable set. Suppose $(U,V)$ were a separation. By [A2] the complements of $U$ and of $V$ are at most countable, and those complements are $V$ and $U$ respectively, so both $U$ and $V$ are nonempty and at most countable with $U\cup V=X$.

Particular point $p$. Every nonempty open set contains $p$ by [A2], so two nonempty open sets meet and $X$ is connected by [A1]; its only component is $X$ by [A3].

In the situation of step 1.5, [A5] gives surjections $f,g:\mathbb{N}\to U$ and $\mathbb{N}\to V$, and $h:\mathbb{N}\times \mathbb{N}\to X$ with $h(0,k):=f(k)$ and $h(m,k):=g(k)$ for $m\mathrm{\ne }0$ is onto $U\cup V=X$; composing with a bijection $\mathbb{N}\to \mathbb{N}\times \mathbb{N}$ from [A5] shows $X$ at most countable, contrary to hypothesis. So the cocountable topology on an uncountable $X$ is connected.

Particular point is locally connected and path-connected. Every nonempty open $U$ contains $p$ and carries as a subspace the particular-point topology on $U$ with the same $p$, hence is connected by step 1.6, so [A3] is witnessed by $U$ itself. For $x,y\in X$ define $\gamma (0):=x$, $\gamma (1):=y$ and $\gamma (t):=p$ for $0<t<1$; the preimage of an open $V$ is $\mathrm{\varnothing }$ if $V=\mathrm{\varnothing }$, and otherwise contains $(0,1)$ and is one of $(0,1)$, $[0,1)$, $(0,1]$, $[0,1]$, all open in $[0,1]$ by [A6]. So $\gamma$ is continuous by [A4] and is a path from $x$ to $y$.

Cofinite and cocountable are locally connected. A nonempty open $U$ carries as a subspace the cofinite, respectively cocountable, topology on $U$ by [A2] and [A1]; $U$ is infinite, respectively uncountable, its complement being finite, respectively at most countable, while $X$ is not. So $U$ is connected by step 1.4, respectively step 2.1, and being open and containing each of its points it witnesses [A3].

Sierpinski space is the particular-point topology on a two-point set with particular point its open point, by [A2], so steps 1.6 and 2.2 apply to it verbatim; and every nonempty connected space has its whole underlying set as its unique component by [A3]. This completes the table.