Every compact subset must be closed? This is an exercise from a topological book.
In $T_1$ space, every compact subset must be closed? For any two compact subset, their intersection must be compact?
Thanks for any help:)
 A: I would like to advertise the following result,   and some  consequences, just in case it is  not so well reported: 
Theorem: Let $B,C$ be compact subsets of $X,Y$ respectively and let $\mathcal W$ be a cover of $B \times C$ by sets open in $X \times Y$. Then $B,C$ have open neighbourhoods $U,V$ respectively such that $U \times V$ is covered by a finite number of sets of $\mathcal W$. 
This has a number of consequences from "the product of two compact spaces is compact" to "a compact subset of a Hausdorff space is closed". (Two others are listed in my book "Topology and groupoids", p. 86, but I forget where I first learned of the theorem, back in the 1960s!) 
A: Claim 1. If $X$ is a compact subset of a Hausdorff space $(Y, \tau)$ then $X$ is closed.
Proof. It sufficies to show that $Y\setminus X$ is open. To
show that $Y\setminus X$ is open, let $y \in Y \setminus X$. For each $x \in X$ fix disjoint $U_x, V_x \in \tau$ so
that $x \in U_x$ and $y \in V_x$. From the open cover $\{U_x : x \in X\}$ of $X$ extract a finite subcover, say $\{U_{x_1},U_{x_2},\ldots,U_{x_n}\}$. Then $V_{x_1}\cap V_{x_1},\cap\ldots,\cap V_{x_n}$ is a neighborhood of $y$ in $Y$ that does not intersect $X$.
Claim 2. If $X$ is compact subset of a topological space $(Y, \tau)$ and $K\subset X$ is a closed subset then $K$ is compact.
Proof. Suppose that $\{V_\alpha: \alpha\in I\}\subset \tau$ is an arbitrary open cover for $K$.
Since $K$ is closed, $Y\setminus K$ is open. Then $X\subset Y$ is covered by $V_\alpha(\alpha\in I)$ and $Y\setminus K$. Since $X$ is compact, it can be covered by a finite number of open subsets $V_{\alpha_1},\ldots, V_{\alpha_n}$ and $Y\setminus K$, and so is $K\subset X$. Hence $K$ is compact. 
Claim 3. If $X_1, X_2$ are closed subsets of a topological space $Y$ then $X_1\cap X_2$ is also closed.      
Claim 4. If $X_1, X_2$ are compact subsets of a Hausdorff space $Y$ then $X_1\cap X_2$ is also compact.      
Counterexample. If $(Y,\tau)$ is not a Hausdorff space then Claim 1. is not valid. Indeed, let $Y=\{a,b\}$ and $\tau=\{\emptyset, X\}$. Note that $Y$ is not a Hausdorff space. Moreover, $\{a\}$ is a compact subset of $Y$ but $\{a\}$ is not closed.
A: Consider the finite complement topology on an infinite set. Is it $T_1$? Which subsets are compact? Which are closed?
A: A space that gives negative answers to both questions is the segment $[-1,1]$ with doubled origin.
More precisely, take two copies $I_0 = [-1,1]\times \{0\}$ and $I_1 = [-1,1] \times \{1\}$ of that interval and identify $(t,0)$ with $(t,1)$ whenever $t \neq 0$, that is, we consider the quotient space $Q$ of $[-1,1] \times \{0,1\}$ modulo the equivalence relation generated by $(t,0) \sim (t,1)$ whenever $t \neq 0$.
The space $Q$ is $T_1$: a set in $Q$ is closed if and only if its pre-image under the quotient map is closed, and a pre-image of a point consists of one or two points. Since continuous images of compact sets are compact, the images $C_i$ of $I_i = [-1,1] \times \{i\}$ under the quotient map are both compact, but they are open and non-closed (look at the pre-images under the quotient map again). The intersection $C_0 \cap C_1$ is homeomorphic to $[-1,1] \smallsetminus \{0\}$, in particular it is non-compact.
Note: In order to obtain an example you need to intersect two non-closed compact sets, otherwise the claims in @blindman's answer show that the intersection is compact.
