Let $X$ and $Y$ be topological spaces such that $Y$ is compact. Then how to show that the projection map $\pi_1 \colon X \times Y \to X$ is a closed map?
My effort:
Let $C$ be a non-empty closed set in $X \times Y$, and let $A \colon= \pi_1 (C)$. We need to show that $X-A$ is open in $X$.
Let $x_0 \in X-A$. Then $x_0 \times Y \subset (X \times Y) -C$. Let $N\colon= (X \times Y ) - C$. Then $N$ is opn in $X \times Y$ and $x_0 \times Y \subset N$.
Since $Y$ is compact, there is an open set $W_0$ in $X$ such that $x_0 \in W_0$ and $W_0 \times Y \subset N$, by the tube lemma (i.e. Lemma 26.8 in Munkres).
Now if $x \in W_0$, then since $W_0 \times Y \subset (X \times Y) - C$, we have $(W_0 \times Y) \cap C = \emptyset$ and so $x \times y \not\in C$ for any $y \in Y$; therefore $x = \pi_1(x\times y) \not\in A = \pi_1(C)$, which shows that $W_0 \subset X-A$, and hence it follows that $X-A$ is open and thus $A = \pi_1(C)$ is closed.
Is the above proof correct and clear enough? If correct, can we make it any clearer?