[proofplan] To prove [path connectedness](/page/Path%20Connectedness), we take two arbitrary points of $X$ and explicitly build a path from one to the other. The path may jump set-theoretically at $0$, but this causes no continuity problem because the codomain has the [indiscrete topology](/page/Indiscrete%20Topology). Indeed, the only open subsets of $X$ are $\varnothing$ and $X$, whose preimages under any map are respectively $\varnothing$ and the whole domain. [/proofplan] [step:Choose arbitrary endpoints and define a path between them] Let $x,y\in X$ be arbitrary. Let $[0,1]$ carry its usual [subspace topology](/page/Subspace%20Topology) inherited from $\mathbb{R}$. Define the function $\gamma:[0,1]\to X$ by setting $\gamma(0)=x$ and $\gamma(t)=y$ for every $t\in(0,1]$. This function has the required endpoint values: \begin{align*} \gamma(0)=x \end{align*} and \begin{align*} \gamma(1)=y. \end{align*} Thus it remains only to prove that $\gamma$ is continuous. [/step] [step:Verify continuity using the indiscrete topology] Let $U\in\tau$ be an arbitrary open subset of $X$. Since $\tau=\{\varnothing,X\}$, either $U=\varnothing$ or $U=X$. If $U=\varnothing$, then $\gamma^{-1}(U)=\varnothing$, which is open in $[0,1]$. If $U=X$, then $\gamma^{-1}(U)=[0,1]$, which is open in $[0,1]$ with its subspace topology. Therefore the preimage under $\gamma$ of every open subset of $(X,\tau)$ is open in $[0,1]$, so $\gamma:[0,1]\to X$ is continuous. [guided] We must check continuity by the open-set definition: for every open subset $U$ of the codomain $(X,\tau)$, the preimage $\gamma^{-1}(U)$ must be open in the domain $[0,1]$. Here the codomain topology is the indiscrete topology, so there are only two possible open sets: \begin{align*} \tau=\{\varnothing,X\}. \end{align*} First suppose $U=\varnothing$. No point of $[0,1]$ maps into $\varnothing$, so \begin{align*} \gamma^{-1}(\varnothing)=\varnothing. \end{align*} The empty set is open in every [topological space](/page/Topological%20Space), in particular in $[0,1]$ with its usual subspace topology. Now suppose $U=X$. Since $\gamma$ has codomain $X$, every point of $[0,1]$ maps into $X$, so \begin{align*} \gamma^{-1}(X)=[0,1]. \end{align*} The whole space is open in every topology, so $[0,1]$ is open in itself. These are the only open subsets of the codomain. Hence the preimage of every open subset of $(X,\tau)$ is open in $[0,1]$, and therefore $\gamma$ is continuous. The apparent jump at $0$ does not matter, because the codomain topology is too coarse to detect it. [/guided] [/step] [step:Conclude that every pair of points is joined by a path] For the arbitrary points $x,y\in X$, we have constructed a continuous map $\gamma:[0,1]\to X$ satisfying $\gamma(0)=x$ and $\gamma(1)=y$. Hence every pair of points of $X$ can be joined by a path. Since $X$ is nonempty by hypothesis, $(X,\tau)$ is path connected. [/step]