Show that $\pi(X, x) = [e_x]$ if $X$ is a finite topological space with the discrete topology. Show that $\pi(X, x) = [e_x]$ if $X$ is a finite topological space with the discrete topology. 
I want to show this by showing that there does not exist any path $f$, $\forall x, y \in X$. Assume for contradiction that there exists a path $f: I \mapsto X$ from $x$ to $y$. Then there are only finitely many points that image of $f$ passes through. Hence, the points that $f$ passes through are open sets whose inverse images under $f$ are disjoint, proper, non-empty open sets on $I$, which then form a separation on $I$. This contradicts that $I$ is connected. So there does not exist any $f$ between distinct points in $X$. Hence $\pi(X, x)$ is trivial. 
Is this a legit proof? Any comments or corrections? Thanks!!! 
 A: The quickest way is to just fix any $x \in X$ and look at a loop i.e a continuous map $f: [0,1] \to X$ with $f(0) = f(1) = x$. It follows that $f([0,1])$ must be connected. Equipping $X$ with the discrete topology says that each singleton is open. We have $f([0,1])  =  \bigcup\limits_{z \in A \subset X} \{z\}$ which is connected only if $A = \{x\}$ i.e every loop is the constant one at $x \in X$. 
A: Your proof is pretty good. However, there are a couple of holes in your contradiction argument.
First, the statement "there does not exist any path $f$, $\forall x,y \in X$" is false, because if $x=y$ then there does exist a path, namely the path which is constant at $x$. So you should instead say "$\forall x,y \in X$, if $x \ne y$ then there does not exist any path $f$ from $x$ to $y$".
Second, your connectivity argument has a gap, because you failed to justify properness of the subsets. You must use the fact that $x \ne y$ to justify that the inverse image under $f$ of each point of $X$ is a proper subset of $I$.
