I'm asked to prove that every totally disconnected topological space satisfies $T_1$ seperation axiom. my proof is this: fix $x\in X$. now, for each $y\in X\setminus\{x\} $ we know that $\left\{ x,y\right\}$ is not connected, therefore there are non-empty disjoint open sets in $X$, say $U,V_{y}$ such that $\left(U\cap\{ x,y\} \right)\cup\left(V_y \cap \{x,y\} \right)=\{x,y\} $. WLOG we can assume that $x\in U, y\in V_{y}$ and we get
$$X\setminus\{x\} =\bigcup_{y\in X\setminus\{x\} } V_y$$
hence every singleton is closed and $X$ is $T_1$
I've seen different proofs and I would like to know if this one is correct.
\backslash
instead of\setminus
, you don't get the horizontal spacing that is proper for a binary operation symbol. I edited accordingly. Thus I changed $X\backslash\{x\}$ to $X\setminus\{x\}.$ (I actually prefer $X\smallsetminus\{x\},$ but that's another matter.) $\endgroup$