Solution.

The image is compact

Let \(f:X\rightarrow Y\) be continuous with \(X\) compact, and let \(\{V_i\}\) be an open cover of \(f(X)\). Each \(f^{-1}(V_i)\) is open by continuity, and these sets cover \(X\): any \(x\) has \(f(x)\in V_i\) for some \(i\). By compactness finitely many suffice, say \(f^{-1}(V_1),\dots ,f^{-1}(V_n)\), and then \(V_1,\dots ,V_n\) cover \(f(X)\).

Bounds attained

Take \(Y=\mathbb {R}\). Then \(f(X)\) is a compact subset of \(\mathbb {R}\), hence closed and bounded by Heine–Borel. Bounded and non-empty, it has a finite supremum \(M\); closed, it contains \(M\). So \(M=f(p)\) for some \(p\in X\), and the infimum likewise.

This is the extreme value theorem, and the proof shows what it really depends on: not differentiability, not even the real numbers, but compactness of the domain and continuity of the map.