= Solution
Suppose $f$ is <continuous> and $x\in\overline A$. Every <open neighbourhood> $V$ of $f(x)$ has an <open> <preimage> containing $x$, which meets $A$. Hence $V$ meets $f(A)$, proving $f(x)\in\overline{f(A)}$.
Conversely, suppose the indicated inclusion holds for every subset. To prove <continuity>, take any <closed set> $C\subseteq Y$ and put $A=f^{-1}(C)$. Since $f(A)\subseteq C$ and $C$ is <closed>,
$$
f(\overline A)\subseteq\overline{f(A)}\subseteq C.
$$
Thus $\overline A\subseteq f^{-1}(C)=A$, so $A$ is <closed>. Every <closed set> has <closed> <preimage>, which is the closed-set characterization of <continuity>.
$$
\boxed{f\text{ continuous}\iff f(\overline A)\subseteq\overline{f(A)}\text{ for every }A\subseteq X.}
$$
This is the <closure characterization of continuity>; it requires no <Hausdorff> or <metric> hypothesis.
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