Let $X, Y$ be Hausdorff spaces. Then $X\times Y$ is a Hausdorff space.
My proof: Let $(x,y)$, $(x',y')$ be distinct points in $X\times Y$. Assume that $x\ne x'$ and $y\ne y'$, then because $X$ and $Y$ are Hausdorff, there exist disjoint neighborhoods $U, U'$ of $x,x'$ and disjoint neighborhoods $V,V'$ of $y,y'$. Then $U\times V$, $U'\times V'$ are disjoint neighborhoods of $(x,y),(x',y')$, so that $X\times Y$ is Hausdorff.
If $x=x'$, then choose disjoint neighborhoods $V,V'$ of $y,y'$. Then $X\times V$, $X\times V'$ are disjoint neighborhoods of $(x,y), (x',y')$. (Symmetrical argument for when $y=y'$.)
Have I missed anything with this proof? I ask because it is simpler than another proof I read.
I don’t even think you need to consider the first case in your argument. As long as one of the coordinates differ you are set. Also, a professor might want you to show that $X\times V$ and $X\times V’$ are open. It is true, in fact they are basic open sets.