Let $G$ be a compact group acting on a compact topological space $X$, that is the function $$G\times X\rightarrow X$$ $$(g,x)\mapsto g\cdot x$$ is continuous and satisfies $$g_1\cdot(g_2\cdot x )=(g_1g_2)\cdot x$$ $$e\cdot x=x$$ for any $g_1,g_2\in G$ and $x\in X$, where $e\in G$ is the neutral element.
Does this imply that the fixed point set of this action, $$ X^G:=\{x\in X|g\cdot x=x \text{ for any $g\in G$}\}, $$ is compact?
Remark. We consider the subset topology on $X^G$.
Assuming $X$ is Hausdorff, yes (and you don't need $G$ to be compact). Since $X$ is Hausdorff, for any $g\in G$, the set $\{x\in X:g\cdot x=x\}$ is closed in $X$. Since $X^G$ is just the intersection of these sets, it is also closed, and hence compact.
(If $X$ is not Hausdorff, then all bets are off. For instance, let $A$ be an infinite set and let $X=A\cup\{x,y\}$, topologized by saying a set is open iff it is either contained in $A$ or is all of $X$. Then $X$ is compact, and a cyclic group $G$ of order $2$ acts on $X$ by swapping $x$ and $y$ and fixing $A$. But then $X^G=A$ is an infinite set with the discrete topology, which is not compact.)