Proof that the continuous image of a compact set is compact

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.

Let any sequence $(y_n) \subset F(K)$ that converges to some $y$. Then by definition there is some $(x_n) \subset K$ such that $F(x_n)=y_n$ for every $n$. Since $K$ is compact (and thus closed and bounded), there is a subsequence $(x_{n_k}) \subset (x_n)$ that converges to some $x\in K$ (see Bolzano Weierstrass Theorem). By continuity of $F$ we have $$ y = \lim_{k \to \infty}y_{n_k} = \lim_{k \to \infty}F(x_{n_k}) = F(x).$$ It follows that $y \in F(K),$ and thus $F(K)$ is closed.

Now suppose that there is a sequence $(w_n) \subset F(K)$ such that $n \leq w_n$ for every $n \in \mathbb{N}$. Again for every $n$ there is $v_n \in K$ such that $F(v_n)=w_n$. $(v_n)$ is contained in the compact $K$ and thus admits a subsequence $v_{n_k}$ converging to some $v \in K$. But this is not possible since $F(v) \in \mathbb{R}$ and $$\infty = \lim_{k \to \infty }n_k \leq \lim_{k \to \infty } F(v_{n_k})=F(v).$$ A contradiction, thus $F(K)$ must be bounded.

It finally follows that $F(K)$ is compact.