 Solution

False: the target need not equal the image. Take the one-point compact space $M=\{0\}$, $N=\mathbb{R}$, and the continuous map $f(0)=0$. The target is not compact, for example because it is unbounded.

The valid theorem is that a continuous image of a compact space is compact. Thus $f(M)$ is compact; compactness of $N$ follows if surjectivity is added.