In this answer
https://math.stackexchange.com/a/2175371/829738
more precisely in this section
"Done that, now prove that $[0, 1]/∼$ is homeomorphic to $\mathbb R/ \mathbb Z$. For that, consider $f:[0, 1]→ \mathbb R$ the inclusion, compose with the quotient to yield a function $[0,1]→\mathbb R/ \mathbb Z$, then use the universal property to get a function $[0,1]/∼→\mathbb R/ \mathbb Z$, which is a continuous bijection from a compact set to a Hausdorff set again."
the map $[0, 1] \rightarrow \mathbb R/ \mathbb Z$ is a quotient map for what reason? Compact-Hausdorff?
There is no claim that the composite map from $[0,1]$ to $\Bbb R/\Bbb Z$ is a quotient map; what matters is that it’s continuous. The goal is to show that $\Bbb R/\Bbb Z$ is compact.
Suppose that $f:X\to Y$ is a continuous bijection from a compact space to a Hausdorff space. If $K\subseteq X$ is closed, then $K$ is compact, so continuity of $f$ ensures that $f[K]$ is compact, and Hausdorffness of $Y$ then ensures that $f[K]$ is closed in $Y$. Thus, $f$ is a closed, continuous bijection and is therefore a homeomorphism, so $\Bbb R/\Bbb Z$ is compact.