= Solution
\b[The claim is false for an arbitrary codomain $Y$.] Let $X=[0,1]$ with its usual topology, let $Y$ be the <Sierpiński space> with open sets $\varnothing,\{1\},Y$, and define $f(0)=0$, $f(x)=1$ for $x>0$. This is a <continuous map>: the only nontrivial open-set preimage is $f^{-1}(\{1\})=(0,1]$, which is relatively open in $X$.
The sets $K_n=[0,1/n]$ are nonempty, nested and compact. But
$$
\boxed{f\!\left(\bigcap_nK_n\right)=\{0\}\ne\{0,1\}=\bigcap_nf(K_n).}
$$
Thus the extra generality in the printed codomain changes the assertion materially. This is the <failure of nested compact image intersection without closed fibers>.
If the codomain is a <T1 space>, the <continuous images of nested compact intersections> theorem does give the desired equality. The forward inclusion holds for every map. For the reverse, take $y\in\bigcap_nf(K_n)$ and set $F_n=K_n\cap f^{-1}(\{y\})$. The fibers are closed by continuity and the T1 property, so the $F_n$ are nested nonempty closed subsets of the compact set $K_1$. The <finite intersection property> gives $x\in\bigcap_nF_n$, with $f(x)=y$. It is enough that these particular fibers be closed in $K_1$; a Hausdorff codomain, as in the native self-map formulation, is a familiar sufficient assumption.
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