 Solution

Use three compactness facts: a closed subset of a compact space is compact; a continuous image of a compact space is compact; and a compact subset of a Hausdorff space is closed. For a closed $C\subseteq X$, these facts show that $f(C)$ is closed in $Y$. Therefore the bijection $f$ is a closed map, and $(f^{-1}{)}^{-1}(C)=f(C)$ proves continuity of its inverse. Thus $f$ is a homeomorphism. This proves the compact-to-Hausdorff continuous bijection theorem without requiring $X$ itself to be Hausdorff.

Let $q:[0,1]\to [0,1]/\sim$ be the quotient map and $h(t)=(\cos 2\pi t,\sin 2\pi t)$. The map $h$ is continuous and identifies precisely the equal parameters and the endpoint pair. Hence it induces a bijection $\bar{h}$ from the quotient space to the unit circle, with $h=\bar{h}\circ q$; the definition of the quotient topology makes $\bar{h}$ continuous. The quotient is compact as a continuous image of $[0,1]$, while the circle is Hausdorff as a subspace of ${\mathbb{R}}^2$. Applying the preceding theorem gives