It is well-known that if $f\colon X\to Y$ is a continuous surjection, $X$ is compact and $Y$ is Hausdorff, then the function $f$ is closed.1 (In particular, if we add the requirement that $f$ is a bijection, we get that $f$ is also a homeomorphism.)
It might be tempting to ask what happens if we change closed map to open map in this result. I.e., we get the question: Is it true that a continuous surjection from a compact space to Hausdorff space is open?
Let me add also a counterexample for this. (I.e., an example where the target space is Hausdorff. This was not required in the original question.) Naturally, to get such an example we need a function which is not bijective. (Otherwise we would get a homeomorphism - which is an open map.)
Let us consider the unit interval $I=[0,1]$ and the unit circle $S=\{(\cos x,\sin x); x\in[0,2\pi]\}$. Then we have a very natural map $f\colon I\to S$
$$f\colon x\mapsto (\cos x,\sin x).$$
This map is surjective and continuous. (In fact, it is a quotient map.) However, it is not open. For example, the set $[0,1/2)$ is open $I$, but the image $f[U]$ is not open in $S$. (No open neighborhood the point $f(0)=(1,0)$ is contained in $f[U]$.)
1For example: Continuous function from a compact space to a Hausdorff space is a closed function