Is this correct definition of $T_1$ space? I found this in a handwritten note:

Defination : A topological space $X$ is $T_1$ if $\forall x \neq y \in X$
  there exist a neighborhood of $y$ such that s.t. $x \not\in V$.

I was almost certain that this was incorrect, because it looked like the definition for $T_0$ space.
However, subsequently this definition was used to prove that all singletons are closed, for example:
$(\Leftarrow)$ Suppose $\{x\}$ is closed, then $\{x\}^c$ is open and $y \in \{x\}^c$ satisfies the definition of a $T_1$ space.
Is this some alternative definition of $T_1$ space?

Yes, this definition is correct.  It is well-known that a space is $T_1$ iff all singleton sets are closed.  For the other direction of the proof, suppose $X$ is $T_1$, let $x\in X$, and let $U$ be the union of all open sets which do not contain $x$.  Then $U$ is open since it is a union of open sets, and does not contain $x$.  Since $X$ is $T_1$, for any $y\neq x$, there is an open $V$ such that $y\in V$ and $x\not\in V$.  The set $V$ is then one of the sets whose union is $U$, so $V\subseteq U$.  In particular, $y\in V$.  Since $y\neq x$ was arbitrary, this means $U=X\setminus\{x\}$.  Thus $X\setminus\{x\}$ is open, so $\{x\}$ is closed.
The difference between this definition and the definition of $T_0$ is that in the definition of $T_0$, all you know is that there exists an open set $V$ such that either $y\in V$ and $x\not\in V$ or $x\in V$ and $y\not\in V$.  So for example, consider the space $S=\{0,1\}$ with topology $\{\emptyset,\{0\},\{0,1\}\}$.  This space is $T_0$: the only pair of points with $x\neq y$ is $0$ and $1$, and $\{0\}$ is a set which contains one but not the other (for some ordering of the two points).  However, this space is not $T_1$, since if you set $x=0$ and $y=1$, there is no open set which contains $y$ but not $x$.
