Consider a topological space $(X, \mathcal{T})$. Suppose $X$ is compact and $(Y, \mathcal{T}_Y)$ is Hausdorff. Let $\Phi: X \times Y \rightarrow Y$ be the projection map. We show that $\Phi$ is a closed mapping. Suppose $A \subset X \times Y$ is closed. Consider an arbitrary $y \notin \Phi(A)$ i.e $(x,y) \notin {A}$ for all $x \in X$. It follows that $X \times {y} \subset (X \times Y) - A$. As $X$ is compact, by the Tube Lemma there exists an open set $V$ of $y$ such that $X \times V \subset (X \times Y) - A$. This implies that $V \subset Y- \Phi(A)$. Since $Y- \Phi(A)$ is open, $\Phi(A)$ is closed.
Is this correct?
Is the Hausdorfness of Y relevant?
Just a couple of comments:
I know that some mathematicians write $X\times y$, but it’s very sloppy; $X\times\{y\}$ is much better.
Where you have ‘an open set $V$ of $y$ such that’, you really mean ‘an open nbhd $V$ of $y$ such that’.
Otherwise it’s correct. And since you did not use Hausdorffness of $Y$ at any point, that means that Hausdorffness of $Y$ is not needed.