If $f : M \to N$ is closed, $\forall\, y \in V \subset N$, $\exists\, U \subset N$ such that $f^{-1}(y) \subset f^{-1}(U) \subset V$ DEFINITION: Let $M$, $N$ be two metric spaces. $f:M\rightarrow N$ is a closed function when for all closed $F\subset M$ its image $f(F)$ is closed in $N$. 
PROBLEM $f:M\rightarrow N$ is a closed function if and only if for all $y\in N$ and all open $V\subset M$ with $f^{-1}(y)\subset V$, there exists an open $U\subset N$ such that $f^{-1}(y)\subset f^{-1}(U) \subset V$.
PROOF ATTEMPT: ($\rightarrow$) Let $f$ be a closed function, $y\in N$ and $V\subset M$ such that $f^{-1}(y)\subset V$. Since V is open and $f$ is closed, $M-V$ is a closed and $f(M-V)$ is closed. Then, $N-f(M-V)$ is open, and since $f^{-1}(y)\cap M-V=\emptyset$, then ${y}\cap f(M-V)=\emptyset$ $\implies$ $y\in N-f(M-V)$ which is open, then there is some open $U$ such that $y\in U\subset N-f(M-V)\implies U\cap f(M-V) = \emptyset \implies U \subset f(V) \implies y\subset U\subset f(V)$ and if we apply $f^{-1}$, then $f^{-1}(y)\subset f^{-1}(U) \subset V$.
QUESTION: Is ($\rightarrow$) correctly proven? How could I prove ($\leftarrow$) ? Thanks so much for your answers.
 A: The following statement $ f^{-1}(y)\cap f(M-V)=\emptyset $ does not make sense because $ f^{-1}(y)$ and $ f(M-V)$ are in different metric spaces. You need to redo this part.
Drawing a graph will help you staying focused in your spaces.  
A: You still need to fix the statement of the problem, but your proof is correct now.  To prove the other direction, suppose that $f$ is not closed, so there is a closed $C\subset M$ such that $f(C)$ is not closed.  Let $y\in \bar{f(C)}\setminus f(C)$.  $f^{-1}(y) \in C^c$ open, so there is a $V\subset N$ open such that $f^{-1}(y) \in f^{-1}(V) \subset C^c$.   However, because $y$ is in the closure of $f(C)$, by definition this means that $V\cap f(C) \neq \emptyset$, so let $z$ be in the intersection.  Since $z\in f(C)$, there must be $x\in C$ such that $f(x)=z$.  This means that $x\in f^{-1}(V)$ since $f(x)\in V$.  But $x\notin  C^c$, which contradicts $f^{-1}(V)\subset C^c$.  
A: So $f: M \to N$ is a closed map iff

$(\ast)$ for every $y \in N$ and every open subset $V$ of $M$ such that
  $f^{-1}[\{y\}] \subseteq V$, there exists an open subset $U$ of $N$ such that $y \in U$ and $f^{-1}[U] \subseteq V$.

You basically already showed that $f$ closed implies $(\ast)$:
Given $y$ and $V$, define $U = N\setminus f[M\setminus V]$, which is open as $M\setminus V$ is closed, hence so is its image and then the complement is open in $N$ again.
We just need to verify two things:
$y \in U$. Suppose it were not. Then $y \in f[M \setminus V]$, so then there is an $x \in M \setminus V$ such that $f(x) = y$. But this $x \in f^{-1}[\{y\}]$ but $x \notin V$, contrary to what was given. So this contradiction shows $y \in U$.
$f^{-1}[U] \subseteq V$. So let $x \in f^{-1}[U]$ be arbitary. 
We need to show that $x \in V$. Suppose it were not. Then $x \in M\setminus V$ and so $f(x) \in f[M \setminus V]$ which by definition of $U$ (as its complement) means $f(x) \notin U$. but this contradicts $x \in f^{-1}[U]$ and this contradiction shows $x \in V$, as required.
Now, for the reverse implication: suppose that $f$ obeys $(\ast)$ and let $C \subseteq M$ be closed. Suppose that $f[C]$ were not closed, then we would have some $y \in \overline{f[C]}$ such that $y \notin f[C]$.
Now note that the latter implies that $f^{-1}[\{y\}] \subseteq M\setminus C$ (if there are points mapping to $y$, they cannot lie in $C$). Let $V = M\setminus C$ (open as $C$ is closed) so that we can apply $(\ast)$ to get an open neighbourhood $V$ of $y$ such that $f^{-1}[U] \subseteq V= M \setminus C$. Now, as $y$ lies in the closure of $f[C]$, there must be some $z \in f[C] \cap U$. So write $z=f(c)$ for some $c \in C$, and observe that $c \in f^{-1}[U]$ (as $z = f(c) \in U$) but $c \notin V =M \setminus C$ (as $c \in C$), and this contradicts how $U$ was chosen. The contradiction shows that $f[C]$ is in fact closed.
