TITLE: Proof that the continuous image of a compact set is compact
QUESTION [15 upvotes]: Let $X\subset \mathbb R^{n}$ be a compact set, and $f :\mathbb R^{n}\to \mathbb R $ a continuous function. Then, $F(X)$ is a compact set.
I know that this question may be a duplicate, but the problem is that I have to prove this using real analysis instead of topology. 
I'm struggling with proving that $F(X)$ is bounded. I know that the image of a continuous function is bounded, but I'm having trouble when it comes to prove this for vectorial functions. 
If somebody could help me with a step-to-step proof, that would be great.

REPLY [13 votes]: Take any open cover of F(K), as F is continuous, the inverse images of those open sets form an open cover of K. Since K is compact there is a finite subcover. By construction, the images of the finite subcover give a finite subcover of F(K), therefore F(K) is compact.