Proof

By hypothesis, $(X,\mathcal O_X)$ and $(Y,\mathcal O_Y)$ are topological spaces and $f:X\to Y$ is a function.

Assume, $f$ is continuous, i.e. the inverse image $f^{-1}[B]$ of every open set $B$ in $Y$ is open in $X.$
In particular, if $B\subseteq Y$ and the complement $B^C$ is open in $Y,$ then $f^{-1}[B^C]$ is open in $X.$
Therefore, if $B$ is closed in $Y,$ then $f^{-1}[B]$ is closed in $X.$

Assume, the inverse image $f^{-1}[B]$ of every closed set $B$ in $Y$ is closed in $X.$
Let $B$ be a closed (but fixed) subset of $Y.$
Since $B$ is closed, it equals its closure $B=B^-.$
Let $A\subseteq X$ be any subset of $X$ with $f[A]=B.$
Thus, $f[A]=f[A]^-.$
By assumption, $f^{-1}[B]=f^{-1}[f[A]]$ is closed in $X.$
Therefore, $A^-\subseteq f^{-1}[f[A]].$
Thus, $f[A^-]\subseteq f[f^{-1}[f[A]]=f[A]=f[A]^ -.$

Assume, the image of the closure of a subset $A\subset X$ is contained in the closure of the image of this subset, formally $f[A^-]\subseteq f[A]^-.$
Take $A=X.$ Then $X=X^-.$
Let $x\in X$ and let $N_{f(x)}$ be a neighborhood $N_{f(x)}$ of $f(x)$ in $Y.$
Since $N_{f(x)}$ is a neighborhood, it contains an open set $O\in O_Y$ with $f(x)\in O\subseteq N_{f(x)}.$
By $(1)$, the inverse image $f^{-1}[O]$ is open in $X.$
Set $N_x:=f^{-1}[O].$
Then $N_x$ is open in $X$ with the image $f[N_x]=O.$
Since $f(x)\in O$, we have $x\in N_x.$
Altogether, we have found a neighborhood $N_x$ of $x$ such that $f[N_x]\subseteq N_{f(x)}$ for any given $x\in X$ and any neighborhood $N_{f(x)}$ of $f(x)$ in $Y.$

Assume, for each $x\in X$ and each neighborhood $N_{f(x)}$ of $f(x)$ there exists a neighborhood $N_x$ of $x$ such that $f[N_x]\subseteq N_{f(x)}.$
Let $B\subseteq Y$ be open in $Y.$ We have to show that $f^{-1}[B]$ is open in $X.$
Take $f(x)\in B.$
Since $B$ is open, there is another open set $O\subseteq Y$ contained in $B$ that contains $f(x).$
In other words, $B$ is a neighborhood of $f(x)$ in $Y.$
By assumption, there exists a neighborhood $N_x$ of $x$ such that $f[N_x]\subseteq B.$
Since $N_x$ is neighborhood, it contains an open set $Q\subseteq X$ containing $x.$
Thus, $Q\subseteq f^{-1}[f[Q]]\subseteq f^{-1}[B].$
Thus, every point $x$ in $f^{-1}[B]$ contains an open set $Q$ contained in $f^{-1}[B],$ making $f^{-1}[B]$ open in $X.$
∎

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Steen, L.A.;Seebach J.A.Jr.: "Counterexamples in Topology", Dover Publications, Inc, 1970
Jänich, Klaus: "Topologie", Springer, 2001, 7th Edition