If $\forall V\subseteq X$ where $x\in \overline V; f(x) \in \overline{f(V)}$, then $f$ is continous in $x$ 
Let $f:(X,\tau_X)\to (Y,\tau_Y)$
Prove: If $\forall V\subseteq X$ where $x\in \overline V; f(x) \in \overline{f(V)}$, then $f$ is continous in $x$.

Could someone verify the following proof?
Proof
If $f$ would not be continous in $x$, then
$$(\exists U \text{ neighbourhood of } f(x))(f^{-1}(U) \text{ is not a neighbourhood of } x) $$
$U$ is a neighbourhood of $f(x)$, then $f(x)$ is an interior point of $U \Rightarrow x \in f^{-1}(U). \qquad \color{red}{(A)}$
However $f^{-1}(U)$ is not a neighbourhood of $x$, thus $x$ is not an interiour point of $f^{-1}(U)$. Or $x\in X\setminus (f^{-1}(U))°$.
Which is equivalent with $x\in \overline{X\setminus f^{-1}(U)}$.
Since $X\setminus f^{-1}(U) = \{x\in X: f(x) \notin U\}$ then $f(X\setminus f^{-1}(U)) = \{f(x)\in f(X): f(x) \notin U\}$.
Then $f(x) \in \overline{f(X\setminus f^{-1}(U))}$ means $f(x)\not \in U$ or $f(x)$ adherent to $f(X\setminus f^{-1}(U))$. We reach $\color{green}{(B)}$ a contradiction with $\color{red}{(A)}$.
Remarks
I think the general idea of the proof is good, but I'm not really sure if the notation at $\color{green}{(B)}$ is correct... How do I write down the contradiction?

Same idea, better put:
Suppose that $U$ is a neighbourhood of $f(x)$ such that $f^{-1}[U]$ is not a neighbourhood of $x$, which indeed means that $x \in \overline{X \setminus f^{-1}[U]}$. So by assumption $f(x) \in \overline{f[X \setminus f^{-1}[U]]}$. As $U$ is a neighbourhood of $f(x)$, $U$ intersects $f[X \setminus f^{-1}[U]]$. So some $y = f(p) \in U$ exists such that $p \in X \setminus f^{-1}[U]$, and the latter means that $f(p) \notin U$, which is a contradiction. 
