Reading through theorem 3.20, Rudin's functional analysis (point (a)).
If $A_1,\ldots, A_n$ are compact convex sets in a topological vector space $X$ then $co(A_1 \cup \ldots \cup A_n)$ is compact.
Reporting the proof with my questions
Let $S$ be the simplex in $\mathbb{R}^n$ consisting of all $s = (s_1,\ldots, s_n)$ with $s_i \geq 0, s_1 + \ldots + s_n = 1$. Put $A = A_1 \times \ldots \times A_n$. Define $f : S \times A \to X$ by $$f(s,a) = s_1 a_1 + \ldots + s_n a_n$$ and put $K = f(S \times A)$ It is clear that $K$ is compact and that $K \subset co (A_1 \cup \ldots \cup A_n)$. We will see that this inclusion is actually an equality.
Why is $K$ compact? First of all since $S$ is compact and each $A_i$ is compact then $S \times A$ is also compact, moreover by it's very definition $f$ is continuous a continuous mapping. I'm not sure if compactness of $S \times A$ and continuity of $f$ suffices to show the compactness of $K$. Another thing I was thinking was to cover $f(S \times A)$ with sets of the form
$$ U_{(s,a)} = f(s,a) + V $$
where $V$ is a neighborhood of $0$ in $K$, and my guess is that with appropriate restrictions of $U_{(s,a)}$ I can work out an appropriate finite subcovers, but It seems a bit complicated, and I'm sure the answer is actually very straighforward
Your real question is this:
Let $f : C \to X$ be a continuous map from a compact topological space $C$ to a topological space $X$. Is $f(C)$ compact?
It is a well-known fact that the answer is yes. See any textbook on topology or make an internet search (e.g. https://planetmath.org/continuousimageofacompactsetiscompact).