Proof
(related to Proposition: Equivalent Notions of Continuous Functions)
By hypothesis, $(X,\mathcal O_X)$ and $(Y,\mathcal O_Y)$ are topological spaces and $f:X\to Y$ is a function.
Ad $(1)\Rightarrow(2)$
 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 $X.$
Ad $(2)\Rightarrow(3)$
 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, $f[A^]\subseteq f[f^{1}[f[A]]=f[A]=f[A]^ .$
Ad $(3)\Rightarrow(4)$
 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 $Y.$
Ad $(4)\Rightarrow(1)$
 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 $X.$
∎
Thank you to the contributors under CC BYSA 4.0!
 Github:

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