Definiton.
$f$ is continuous at $a$ provided that for each open set $V$ in $Y$ containing $f(a)$ there is an open set $U$ in $X$ containing $a$ such that $f(U) \subset V$.
Problem.
$f$ is continuous at $a$ iff for each subset $A$ of $X$ with $a\in \overline{A}$, $f(a)\in \overline{ f(A)}$.
The condition $f(\bar A)\subseteq \overline{f(A)}$ for each $A\subseteq X$ can be taken as the definition of continuity (provided one defines a closure operator before one defines open sets, of course). Let's assume the
DEF $f:(X,\mathscr T)\to(Y,\mathscr U)$ is continuous if for each open $A\in Y$, $f^{-1}(A)$ is open in $X$.
We know that this is equivalent to
DEF* $f:(X,\mathscr T)\to(Y,\mathscr U)$ is continuous if for each closed $F\in Y$, $f^{-1}(F)$ is closed in $X$.
Then we have the
THM $f:(X,\mathscr T)\to (Y,\mathscr U)$ is continuous if and only if for each $A\subseteq X$, $f(\bar A)\subseteq \overline{f(A)}$
PROOF Suppose that $f$ is continuous. Given a subset $A$, we have $f(A)\subseteq \overline{f(A)}$, which means $A\subseteq f^{-1}f(A)\subseteq f^{-1}\overline{f(A)}$. This last set is closed, so $\bar A\subseteq f^{-1}\overline{f(A)}$, which means $f(\bar A)\subseteq \overline{f(A)}$.
Conversely, suppose that $f(\bar A)\subseteq \overline{f(A)}$ for each $A\subseteq X$. Let $F$ be closed in $Y$. Then $f(\overline{f^{-1}(F)})\subseteq \overline{f({f^{-1}(F)})}\subseteq \bar F=F $. Thus $\overline{f^{-1}(F)}\subseteq f^{-1}(F)$. Since the converse always holds, we have $f^{-1}(F)=\overline {f^{-1}(F)}$, so this set is closed and $f$ is continuous.
Suppose $f$ is continuous at $a$, according to the definition.
Let $A$ be a subset with $a \in \overline{A}$, we want to show that $f(a) \in \overline{f[A]}$.
We will use that a point is in the closure of a set iff every open neighbourhood of the point intersects that set.
So let $V$ be an open neighbourhood of $f(a)$. Using the definition, we find a neighbourhood $U$ of $a$ such that $f[U] \subset V$. As $a \in \overline{A}$ we know that $V$ intersects $A$, say $a' \in A \cap V$. But then $f(a') \in V \cap f[A]$ (as $f[U] \subset V$), so indeed $V$ intersects $f[A]$, and as $V$ is arbitrary, $f(a) \in \overline{f[A]}$. This shows one implication.
Now let $f$ fulfill the closure condition at $a$; we want to see that $f$ is continuous at $a$. So let $V$ be an open neighbourhood of $f(a)$. Suppose now (striving for a contradiction) that for all open neighbourhoods $U$ of $a$ we have that $f[U] \not\subset V$. This means that every neighbourhood of $a$ contains points of $X \setminus f^{-1}[V]$, so $a \in \overline{X \setminus f^{-1}[V]}$, and so, by the closure condition, $f(a) \in \overline{f[X \setminus f^{-1}[V]} \subset \overline{Y \setminus V}$, but this is false, as $V$ does not intersect $Y \setminus V$. This contradiction shows the continuity of $f$ at $a$.
$\to)$ Hint: Let $A\subseteq X$ and $x\in \overline{A}$ and $V$ be a neighborhood of $f(x)$. It's enough to show $f(A)\cap U\ne \emptyset$. $f^{-1}(U)$ is a neighborhood of $x$. There's some $$z\in f^{-1}(U)\cap A$$ $$\Rightarrow f(z)\in U\cap f(A)$$
$\leftarrow)$ Hint: Let $x\in X$ and $V$ be an open neighborhood of $f(x)$. It's enough to show $f^{-1}(V)$ is a neighborhood of $x$. Suppose it is not. $x$ is in the boundary of $f^{-1}(V)$. So $$x\in \overline{\big(f^{-1}(V) \big)^c}$$ $$\Rightarrow f(x)\in f\big(\overline{\big(f^{-1}(V) \big)^c}\big)\subseteq \overline{f\big(\big(f^{-1}(V) \big)^c\big)}=\overline{f\big(f^{-1}(V^c) \big)}\subseteq \overline{V^c}$$ So $f(x)$ is in the boundary of $V$. A contradiction.
