Intersection of finite number of compact sets is compact?

For Hausdorff spaces your statement is true, since compact sets in a Hausdorff space must be closed and a closed subset of a compact set is compact. In fact, in this case, the intersection of any family of compact sets is compact (by the same argument). However, in general it is false.

Take $\mathbb{N}$ with the discrete topology and add in two more points $x_1$ and $x_2$. Declare that the only open sets containing $x_i$ to be $\{x_i\}\cup \mathbb{N}$ and $\{x_1 , x_2\}\cup \mathbb{N}$. (If you can't see it immediately, check this gives a topology on $\{x_1 , x_2\}\cup \mathbb{N}$).

Now $\{x_i\}\cup \mathbb{N}$ is compact for $i=1,2$, since any open cover must contain $\{x_i\}\cup \mathbb{N}$ (it is the only open set containing $x_i$). However, their intersection, $\mathbb{N}$, is infinite and discrete, so definitely not compact.

I would like to supply another answer to this interesting question. I learnt the following example in a masters thesis of Cecil Eugene Denney (Kansas State University, 1966). Here is a link to the thesis. What follows below is a modified version of Example 3 in page 27 of the thesis.

Example. Let $X=\{a, b\}$ be a topological space whose only open sets are $\emptyset$ and $X$ itself (so $X$ has the indiscrete topology). Let $Y=\mathbb{R}$ be the set of all real numbers with the standard topology. We will consider the topological space $X\times Y$. Consider the following two subsets of $X\times Y$: $$ C = \{(a, y): 1\leq y < 3\} \cup \{(b, y): 3\leq y \leq 4\} $$ and $$ D = \{(a, y): 2<y \leq 4\} \cup \{(b, y): 1\leq y \leq 2\}. $$ We claim that $C$ and $D$ are both compact. Indeed, in given any open cover $\mathcal{U}$ for $C$, each open set must be of the form $X\times U$ where $U$ is an open subset of $Y$ (this is where we use the fact that the only open subsets of $X$ are $\emptyset$ and $X$). Using the projection $\pi_Y: X\times Y \to Y$, we note that the collection of all $U$ such that $X\times U\in\mathcal{U}$ forms an open cover of $\pi_Y(C) = [1, 4]$ which is compact. Thus, there are finitely many $U_i$ which cover $[1, 4]$, and consequently finitely many $X\times U_i$ which cover $C$. This shows $C$ is compact, and the verification for compactness of $D$ is identical.

However, $C\cap D = \{(a, y): 2 < y < 3\}$ is homeomorphic, via the projection $\pi_Y$, to the open interval $(2, 3)$ in $Y=\mathbb{R}$, and it is well-known that $(2, 3)$ is not compact.

In a $T_2$ space (Hausdorff), compact subsets are closed. Let be a $T_2$ space and let $F$ be a family of compact subsets of $S$. To avoid trivial cases let $F$ have at least one non-empty member $f$.

Now $G=\cap F$ is closed and $G^c=S\backslash G$ is open. If $V$ is a family of open sets with $\cup V\supset G,$ then $V'=V\cup \{G^c\}$ is an open cover of $f.$ (Indeed, $\cup V'=S.$)

Since $f$ is compact, there exists a finite $H\subset V'$ with $\cup H\supset f.$ Then $H'=H\backslash \{G^c\}$ is a finite subset of $V$, and $\cup H'\supset G.$

So $G$ is compact.

Note that we needed each member of $F$ to be closed, so that $G$ is closed, so that $G^c$ is open, so that $V'$ is an open family. As an earlier answer shows, if $S$ is not $T_2$ then this can fail.