Closed mapping in Hausdorff space. Consider $(X, \mathcal{T})$. Suppose $X$ is compact, $(Y, \mathcal{T}_Y)$ is Hausdorff. How do we show that the projection map $f: X \times Y \rightarrow Y$ is a closed mapping.
 A: Hausdorffness is not needed, in neither space. 
Helpful fact: $f: X \rightarrow Y$ is a closed map iff for every $y \in Y$ and every open set $O$ where $f^{-1}[\{y\}] \subset O$, there is some open subset $V \subset Y$ that contains $y$ and such that $f^{-1}[V] \subset O$.
Now if $X$ is compact and $Y$ is any space and $f(x,y) = y$ is the projection onto the non-compact space, then let $y \in Y$ and let $O$ be an open set that sits around $X \times \{y\} = f^{-1}[\{y\}]$. Then we find for each $x \in X$, (so $(x,y) \in O$) a basic open set $U_x \times V_x$ of $X \times Y$ such that $x \in U_x, y \in V_x, U_x \times V_x \subset O$. Finitely many of the $U_x$ cover $X$, say $U_{x_1},\ldots,U_{x_k}$; now define $V = \cap_{i=1}^k V_{x_i}$, which is open as a finite intersection of open sets, then $X \times V = f^{-1}[V] \subset O$: let $(x,v) \in X \times V$, then $x$ lies in some $U_{x_i}$, and then $v$ lies in the corresponding $V_{x_i}$ (as it is in $V$, their intersection) and so $(x,v) \in O$ by construction. So our lemma allows us to conclude that $f$ is a closed map.
Proof of our lemma: (so $X$,$Y$ general spaces again). Suppose $f$ satisfies the condition on the fibres. Let $C \subset X$ be closed. Let $y$ be in the closure of $f[C]$, we want to show it is in $f[C]$, so suppose it is not. Then $f^{-1}[\{y\}] \subset X \setminus C$ by assumption, and so we have some $V \subset Y$ open with $y \in V$ and $f^{-1}[V] \subset X \setminus C$. But $V$ intersects $f[C]$, as $y$ is in its closure, so for some $x \in C, f(x) \in V$, but now we have a contradiction as $x \in f^{-1}[V]$ and in $C$ as well, while it should be in $X \setminus C$.
The reverse is also true: suppose $f$ is closed map and $f^{-1}[\{y\}] \subset O$ for some open $O$, then $X\setminus O$ is closed, and so $V = Y \setminus f[X \setminus O]$ is open and contains $y$ and $f^{-1}[V] \subset O$ (nice exercise for you; we only need the one direction I did in detail anyway).
Bonus fact: this property of compact spaces actually characterises compactness:
Suppose that $X$ is a space such that for all spaces $Y$ the projection onto $Y$ from $X \times Y$ is a closed map, then $X$ is compact.
