Under what condition only does every compact subset of $X$ is closed implies $X$ Hausdorff? It is trivial to see that:
If $X$ is Hausdorff, then every compact subset of $X$ is closed.
I am asking under what condition does the converse hold, i.e. when does
If every compact subset of $X$ is closed, then $X$ is Hausdorff
hold.
EDIT:
Sorry for whoever just read the question, I corrected the title and the whole question. It should be clear now.
 A: As for me, this questions sounds unnaturally, but seems to be not so trivial. It seems the following. 
Let $X$ be a space such that  every compact subset of $X$  is closed. Then $X$ is $T_1$, because each  one-point subset is compact.  
Proposition 1. If $X^2$ is a sequential space, then $X$ is Hausdorff. 
Proof. Suppose the opposite. Then the diagonal $\Delta=\{(x,x)\in X^2:x\in X\}$ is not closed in $X^2$. Since $X^2$ is a sequential space then there exists a sequence $\{(x_n,x_n)\}$ of points of $\Delta$, converging to a point $(x,y)\in X^2\backslash\Delta$. Without loss of generality we may suppose that $x_n\not=y$ for each $n$. The sequence $\{x_n\}$ converges to $x$. Therefore a set $X_0=\{x\}\cup\{x_n\}$ is compact. Hence $X\backslash X_0$ is an open neighborhood of $y$. Since the sequence $\{x_n\}$ converges to $y$,
there exists a number $n$  such that $x_n\in X\backslash X_0,$ a contradiction. $\square$
Example 1. There exists a non-Hausdorff space $X$  such that each compact subset of $X$  is closed. 
Put $X=\omega\cup\{\alpha\}\cup\{\alpha’\}$, where $\alpha\not=\alpha’$ and $\{\alpha, \alpha’\}\cap\omega=\varnothing$. Let $\mathcal F$ be a free ultrafilter on the set $\omega$. Define a topology $\tau$ on the $X$ as follows. A subset $U$ of $X$ belongs to $\tau$ iff $U\subset\omega$ or $U\cap\omega\in\mathcal F$. Since $\mathcal F$ is an ultrafilter, we can easily check that each compact subset of $X$ is finite and, hence, closed in $T_1$-space $X$.
The following questions are already answered by Martin. 
For the sake of future advances in this direction I formulate the following  questions. 
Exists a non-Hausdorff space $X$ such that each compact subset of $X$ is closed, provided:


*

*$X$ is compact?

*$X$ is locally compact?

*$X$ is a "$k$-space"?

*$X$ is a "$k_\omega$-space"?

*$X$ is a Fréchet-Urysohn space?

*$X$ is a sequential space?
