Equivalence of Definitions of Connected Topological Space/No Separation iff No Union of Closed Sets

Theorem
Let $T = \left({S, \tau}\right)$ be a non-empty topological space.

Then:
 * $T$ admits no partition

iff:
 * $T$ has no two disjoint non-empty closed sets whose union is $S$.

Proof
Let $T = \left({S, \tau}\right)$ admit no partition.

That is:
 * $\nexists A, B \in \tau: A, B \ne \varnothing, A \cup B = S, A \cap B = \varnothing$

Aiming for a contradiction, suppose $A, B \subseteq S$ are non-empty closed sets of $T$ such that $A \cap B = \varnothing$ and $A \cup B = S$.

Then $A = T \setminus B$ and $B = T \setminus A$ are both open sets which form a (set) partition of $S$.

From this contradiction it follows that $T$ has no two disjoint non-empty closed sets whose union is $S$.

Now let $T$ have no two disjoint non-empty closed sets whose union is $S$.

Aiming for a contradiction, suppose $A, B \in \tau$ form a partition of $T$.

That is:
 * $A \cap B = \varnothing$
 * $A \cup B = S$

Then $A = T \setminus B$ and $B = T \setminus A$ are two disjoint non-empty closed sets whose union is $S$.

From this contradiction it follows that $T$ admits no partition.