= Solution
Use the <equivalence relation> of belonging to the same fibre:
$$
\boxed{x\mathcal R x'\iff f(x)=f(x').}
$$
The map $\bar f:X/\mathcal R\to Y$, $[x]\mapsto f(x)$, is well defined, injective by the definition of the relation, and surjective because $f$ is surjective. It is continuous by the <quotient topology>: for open $V\subseteq Y$, $q^{-1}(\bar f^{-1}(V))=f^{-1}(V)$ is open.
The <quotient space> is <compact> as the continuous image of <compact> $X$, and $Y$ is <Hausdorff>. The <compact-to-Hausdorff continuous bijection theorem> makes $\bar f$ a <homeomorphism>. Indeed, images of closed subsets of a <compact> space are <compact> and hence closed in a <Hausdorff> space, so the inverse is continuous.
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