Equivalence of Definitions of Totally Separated Space

Proof
Let $T = \left({S, \tau}\right)$ be totally separated by definition 1:
 * For every $x, y \in S: x \ne y$ there exists a separation $U \mid V$ of $T$ such that $x \in U, y \in V$.

there exist $x, y \in S$ such that $x \ne y$ such that $x$ and $y$ are in the same quasicomponent of $T$.

Then, by definition of quasicomponent, every separation of $T$ includes a single open set $U \in \tau$ which contains both $x$ and $y$.

But this contradicts our stipulation that there exists a separation $U \mid V$ of $T$ such that $x \in U, y \in V$.

So there exists no quasicomponent of $T$ with more than one element.

That is, $T = \left({S, \tau}\right)$ is totally separated by definition 2.

Let $T = \left({S, \tau}\right)$ be totally separated by definition 2:
 * Each of its quasicomponents is a singleton set.

there exist $x, y$ such that there exists no separation $U \mid V$ of $T$ such that $x \in U, y \in V$.

That is, each separation of $T$ contains an open set $U \in \tau$ which contains both $x$ and $y$.

That is, there exists a quasicomponent which contains both $x$ and $y$.

But this contradicts our stipulation that every quasicomponent of $T$ is a singleton.

So for every $x, y \in S: x \ne y$ there exists a separation $U \mid V$ of $T$ such that $x \in U, y \in V$.

That is, $T = \left({S, \tau}\right)$ is totally separated by definition 1.