Let $p:Y \mapsto X$ be a finite covering map, and let $\sim$ be the equivalence relation on $Y$ induced by $p$: $y_1 \sim y_2$ iff $p(y_1)=p(y_2)$.
By definition, $p$ descends to the quotient, i.e. we have an induced continuous map: $\tilde p:Y/\sim\, \, \, \mapsto \, X$. Clearly $\tilde p$ is bijective. Must it be a homeomorphism?
I am fine with assuming that $X$ is Hausdorff, but not that $Y$ is compact. (If $Y$ were compact, then $Y/\sim$ would also be compact, hence $\tilde p$ would be a continuous bijection from compact to Hausdroff, thus a homeomorphism).
I think that if $p$ is not finite, then $\tilde p$ does not need to be a homeomorphism: Take e.g. $Y=\mathbb{R},X=\mathbb{S}^1$, and $p(t)=e^{it}$. Then $Y/\sim \,\, \cong \,[0,2\pi)$ which is not homeomorphic to $\mathbb{S}^1$.
It is always true, even for nonfinite coverings. Your counterexample is not one, as $Y/\sim$ is then $[0,2\pi]$ with $0\sim 2\pi$, that is, a circle.
The proof is quite easy : $\tilde p$ is quite clearly continuous and bijective as you pointed out. We want to prove that its inverse, $h$ is also continuous.
Continuity is local : let $x\in X$ and let $U$ be a small open set containing $x$ such that $p^{-1}(U)\cong U\times I$ (where $I$ is discrete) as spaces over $U$.
Then the restriction of $h$ to $U$ factors as $U\to Y\to Y/\sim$, where the first map is $u\mapsto (u,i)$ for some fixed $i\in I$. Both these maps are continuous, so $h_{\mid U}$ is continuous as well, so $h$ is continuous.