= Solution
Let $(V_j)_{j\in J}$ be an <open cover> of $f(K)$ by <open subsets> of $Y$. The <continuity> of $f$ makes each <preimage> $f^{-1}(V_j)$ <open> in $X$, and these <preimages> cover $K$. By <compactness>, choose $j_1,\ldots,j_m$ with
$$
K\subseteq\bigcup_{r=1}^m f^{-1}(V_{j_r}).
$$
Applying $f$ gives $f(K)\subseteq\bigcup_{r=1}^m V_{j_r}$. Thus every <open cover> of $f(K)$ has a <finite subcover>, proving the <continuous image of a compact space> theorem. A cover by sets <open> in $f(K)$ gives the same argument after expressing them as intersections with ambient <open sets>.
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