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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. ButThus 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 $y\in {\bigcap }_{n}f(K_n)$ 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 $x\in {\bigcap }_n{F}_n$, 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.