It is a theorem that if $f:X \to Y$ is a continuous bijection, $X$ is compact, then $g = f^{-1}$ is continuous. My professor asked us to find a counterexample to
If $f:X \to Y$ is continuous, $Y$ is compact, then $g = f^{-1}$ is continuous.
I do not like my counterexample so much because it uses the discrete metric. Are there other counterexamples?
My Counterexample: Let $X = [0, 1]$ with the Discrete metric, $Y = [0, 1]$ with the Euclidean metric, and let $f$ be the identity function.
Your example is fine. Another example is to take $X=[0,2\pi)$ with the usual metric, and let $Y=S^1 = \{z\in\mathbb{C}\mid |z|=1\}$, the unit circle on the complex plane, with the usual metric (so the distance is measured on the plane, not along the circle).
Let $f\colon X\to Y$ be given by $f(x) =e^{ix} = \cos(x) + i\sin(x)$.
Now, $Y$ is closed and bounded, so it is compact. The function $f$ is a continuous bijection. But $f^{-1}$ is not continuous; if it were, then $f$ would be a homeomorphism, but removing a single point from $X$ (other than $0$) will disconnect $X$, and removing a single point from $Y$ does not. Alternatively, note that you cannot find a neighborhood of $1\in Y$ whose image lies inside the neighborhood $[0,\frac{1}{2})$ of $f^{-1}(1)=0$, since the image will contain points arbitrarily close to $2\pi$.