Topological space (X,T), where $$X=\{v_1,v_2,e_1,e_2,u_1,u_2\}$$ $$T=\{\emptyset,\{u_1\},\{u_2\},\{u_1,u_2\},\{e_1,u_1,u_2\},\{e_2,u_1,u_2\},\{e_1,e_2,u_1,u_2\},\{v_1,e_1,e_2,u_1,u_2\},\{v_2,e_1,e_2,u_1,u_2\},X\}$$ The question is to prove that the space is path connected. To show that, do I define a function from $[0,1]->X$ such that $f([0,1))=u_1$ and $f(1)=\text{any other elements in }X$? And then show it is continuous? It doesn't seem right though.
I'm also not sure why the topology T is the way it is.
So, if we define $f([0,1))= u_1$ and $f(1) = y$, where $y \neq u_2$, then note that $f^{-1}(\{u_1\})$ is open, and any open set that contains $y$ also contains $u_1$, so for any such set $S$, $f^{-1}(S) = [0,1]$, which is open in $[0,1]$ obviously. Therefore, $f$ is continuous.
A similar logic applies for $u_2$ above.
So every point is path connected to $u_1$ and $u_2$, and you can now use the fact that paths can be reversed and concatenated to finish the argument. For example, to go from $v_1 \to v_2$, first go to $u_1$ via the reverse of the path we made above, and then go to $v_2$ as usual.
(the next part is only for information. If you do not understand, it is fine)
I am inclined to feel that this topology comes from a topology we naturally establish on a poset, by considering chains. Here, we would have $u_i \leq e_i \leq v_i$ for each $i=1,2$, and the chains of this poset form a topology, that I think we call the Alexandroff topology on the poset. I would be inclined to think that the Alexandroff topology is path connected, under certain conditions on the poset.
Indeed, due to local path connectedness, one sees that the Alexandroff topology is path connected if and only if it is connected i.e. there is no partition of the set into disjoint open sets. Since every pair of non-empty open sets intersect at $u_1$ in this case, we see that the Alexandroff topology is connected, and therefore path connected without having to explicitly show the path.