Totally Disconnected Space is Totally Pathwise Disconnected

Theorem
Let $T = \left({S, \tau}\right)$ be a topological space which is totally disconnected.

Then $T$ is a totally pathwise disconnected space.

Proof
Let $T = \left({S, \tau}\right)$ be a topological space which is totally disconnected.

Then by definition $T$ contains no non-degenerate connected sets.

$T$ is not a totally pathwise disconnected space.

That is, there exist two points $x, y \in S$ such that there exists a path between $x$ and $y$.

That is, $x$ and $y$ are path-connected.

Thus they are in the same path component.

From Path-Connected Space is Connected, $x$ and $y$ are therefore in the same component.

But that contradicts the definition of $T$ having no non-degenerate connected sets.

Hence, by Proof by Contradiction, $T$ is a totally pathwise disconnected space.