Let $X=\{(x,0)\in \mathbb{R}^2\,|\,x \in \mathbb{R}\} \cup \{(0,1)\}$, give $(0,1)$ the name "$a$", and let $T=\{A\subseteq X\,|\,a \in A\} \cup \{\emptyset\}$. Is $(X,T)$ path-connected?
I believe it is, and here is why:
Let $f:[0,1] \to X$ be such that $f(0)=x$ and $f(1)=a$, where $x \in X\setminus \{a\}$ (of course the constant map to $a$ is continuous). Then $f([0,1])=\{a\}\cup X_0$ where $X_0$ is closed (does not contain $a$). Then we have $f^{-1}(\{a\}\cup X_0) = f^{-1}(\{a\}) \cup f^{-1}(X_0)$. Since $\{a\}$ is open we must have $f^{-1}(\{a\})$ of the form $(a_0,1]$ (this is open in the subspace topology). Since $X_0$ is closed we must then have $f^{-1}(X_0) = [0,a_0]$. So $f^{-1}(\{a\} \cup X_0) = [0,1]$. I don't see any contradiction in this, so is it true that $X$ is path-connected?
I worked this out because intuitively the space seems not to be path-connected but is due to the topology (if I'm correct). Is there a non-discrete topology on $X$ that makes it non-path-connected?