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?

The definition is correct, albeit a little sloppy. Here’s a slightly more careful version of it:

$X$ is $T_1$ if for each $x\in X$ and each $y\in X\setminus\{x\}$ there is a nbhd $V$ of $y$ such that $x\notin V$.

Let $p$ and $q$ be any two distinct points of $X$. If we set $x=p$ and $y=q$, we see that the definition ensures that $q$ has a nbhd that does not contain $p$. We can just as well set $x=q$ and $y=p$, however, and conclude that $q$ has a nbhd that does not contain $p$. Thus, the definition really does say that $X$ is $T_1$. The point is that since both quantifiers are universal, we can interchange the rôles of $x$ and $y$ in the definition. 
The real problem with the quoted version is that it doesn’t make the quantifier on $x$ explicit. Because of this, it’s easy on first reading to get the impression that the definition is asymmetric in $x$ and $y$, like the definition of $T_0$ separation.
