A map is continuous if and only if for every set, the image of closure is contained in the closure of image

Because the property is stated in terms of closures, it's I think slightly easier to use the inverse image of closed is closed equivalence of continuity instead:

If $C$ is closed in $Y$, we need to show that $D = f^{-1}[C]$ is closed in $X$.

Now using our closure property for $D$: $$f[\overline{D}] \subseteq \overline{f[D]} = \overline{f[f^{-1}[C]]} \subseteq \overline{C} = C,$$ as $C$ is closed.

This means that $\overline{D} \subseteq f^{-1}[C] = D$, making $D$ closed, as required.

If $f$ is continuous, then $f^{-1}(Y-\overline{f(A)})$ is open; since $f^{-1}(Y-\overline{f(A)}) = X -f^{-1}(\overline{f(A)})$, then $f^{-1}(\overline{f(A)})$ is closed; since $A\subseteq f^{-1}(f(A))\subseteq f^{-1}(\overline{f(A)})$, we conclude that $\overline{A}\subseteq f^{-1}(\overline{f(A)})$, and therefore $f\left(\overline{A}\right)\subseteq f\left(f^{-1}\left(\overline{f(A)}\right)\right)\subseteq \overline{f(A)}$. This proves one direction. (Since you said you already had this, I posted the full argument).

Conversely, assume that for every $A$, $f\left(\overline{A}\right)\subseteq \overline{f(A)}$. Let $V$ be an open subset of $Y$; we want to show that $f^{-1}(V)$ is open; equivalently, we want to show that $X-f^{-1}(V)$ is closed; that is, we want to show that $X-f^{-1}(V)=\overline{X-f^{-1}(V)}$.

By assumption, $f\left(\overline{X-f^{-1}(V)}\right)\subseteq \overline{f(X-f^{-1}(V))}$. Now, $X-f^{-1}(V) = f^{-1}(Y-V)$, so $f(X-f^{-1}(V)) \subseteq Y-V$, which is closed; so $$f\left(\overline{X-f^{-1}(V)}\right)=f\left(\overline{f^{-1}(Y-V)}\right)\subseteq\overline{f\left(f^{-1}(Y-V)\right)}\subseteq Y-V.$$ Therefore, $$\overline{X-f^{-1}(V)} \subseteq f^{-1}(Y-V).$$ Is this sufficient?

Henno’s argument is probably the slickest, but one can also work pointwise, either directly or, if one already has those characterizations of continuity, with nets or filters.

Suppose that $f$ is not continuous. Then there is some point $x_0\in X$ at which $f$ fails to be continuous. If $y_0=f(x_0)$, this means that there is an open nbhd $U$ of $y_0$ such that if $V$ is an open nbhd of $x_0$, there is a point $x_V\in V$ such that $f(x_V)\notin U$. Let $$A=\{x_V:V\text{ is an open nbhd of }x_0\}\;.$$

Clearly $x_0\in\overline{A}$, and $f[A]\subseteq Y\setminus U$. Moreover, $Y\setminus U$ is closed, so $\overline{f[A]}\subseteq Y\setminus U$, and hence

$$y_0\in f[\overline{A}]\subseteq\overline{f[A]}\subseteq Y\setminus U\;,$$

contradicting the choice of $U$.

Those who like nets may notice that $A$ actually is a net, over the directed set $\langle\mathscr{N}(x_0),\supseteq\rangle$, where $\mathscr{N}(x_0)$ is the family of open nbhds of $x_0$, and the argument shows that $f$ doesn’t preserve the limit of this net. (I could of course just as well have used the nbhd filter at $x_0$, instead of using only open nbhds.)

Those who prefer filters may modify this to consider the filter $$\mathscr{F}=\{V\cap f^{-1}[Y\setminus U]:V\in\mathscr{N}(x_0)\}$$ instead.