= Solution
<Continuity> means that the <inverse image> of every open subset of $Y$ is open in $X$. If $x\in\overline A$ and $V$ is an <open neighborhood> of $f(x)$, then $f^{-1}(V)$ is an <open neighborhood> of $x$, so it meets $A$. Thus $V$ meets $f(A)$. The <closure> characterization gives
$$
\boxed{f(\overline A)\subseteq\overline{f(A)}.}
$$
If $B$ is empty the assertion is immediate. Otherwise, if $B$ is <bounded>, choose a centre $p$ and $R<\infty$ with $d(p,b)\leq R$ for all $b\in B$. The <closed ball> $\{y:d(p,y)\leq R\}$ is closed because $|d(p,y)-d(p,z)|\leq d(y,z)$. It therefore contains $\overline B$. Hence \b[the <closure> of a <bounded set> is <bounded>]. Empty sets cause no exception. If $f(A)$ is <bounded>, its <closure> is <bounded> by this result, and the displayed inclusion makes \b[$f(\overline A)$ <bounded>] too.
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