= Solution
\b[True.] For any <continuous map>, $f(\overline A)\subseteq\overline{f(A)}$: every open neighbourhood of $f(x)$ pulls back to an open neighbourhood of $x\in\overline A$, and therefore meets $A$ in its preimage.
Apply the same inclusion to the <continuous> inverse of the <homeomorphism>:
$$
f^{-1}(\overline{f(A)})\subseteq\overline{f^{-1}(f(A))}=\overline A.
$$
Applying $f$ gives the reverse inclusion, proving
$$
\boxed{\overline{f(A)}=f(\overline A).}
$$
The closures are taken in their respective ambient spaces.
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