= Solution
The <closure> $\operatorname{cl}(A)$ is the intersection of all <closed subsets> containing $A$; equivalently it consists of the points whose every open <neighbourhood> meets $A$.
Suppose $f$ is continuous and $x\in\operatorname{cl}(A)$. For any open <neighbourhood> $V$ of $f(x)$, <continuity> makes $f^{-1}(V)$ an open <neighbourhood> of $x$, so it meets $A$. Hence $V$ meets $f(A)$, showing $f(x)\in\operatorname{cl}(f(A))$. This proves $f(\operatorname{cl}(A))\subseteq\operatorname{cl}(f(A))$.
Conversely suppose this inclusion holds for every subset $A$. For a <closed subset> $C\subseteq Y$, let $A=f^{-1}(C)$. Then $f(A)\subseteq C$ and therefore $\operatorname{cl}(f(A))\subseteq C$. By the assumed inclusion, $f(\operatorname{cl}(A))\subseteq C$, so $\operatorname{cl}(A)\subseteq f^{-1}(C)=A$. Since $A\subseteq\operatorname{cl}(A)$ always, $A$ is closed. Thus every <closed set> has closed <preimage>, equivalently every <open set> has open <preimage>. This is <continuity>, proving \b[the claimed equivalence].
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