Determine the points of continuity of $h(x)=\lfloor \sin(x) \rfloor$, where $\lfloor x \rfloor$ is the greatest $m \in \mathbb Z$ such that $m \leq x$ (floor function).
This is problem 5.1.4(c) of Bartle and Sherbert Intro to Real Analysis.
Here is my attempt at a solution.
Claim. The function $h$ is continuous on $\mathbb R \setminus \left( \{2\pi n : n \in \mathbb Z\} \cup \{ \pi + 2\pi n: n \in \mathbb Z\} \cup \{\pi / 2 + 2\pi n : n \in \mathbb Z\}\right)$.
Proof. For values of the form $2\pi n$ (for some integer $n$), we have that $\lim_{x\to 2\pi n^-}h(x)=-1$ but $\lim_{x\to 2\pi n^+}h(x)=0$, so $h$ is not continuous at $2\pi n$. Similarly for values of the form $\pi + 2\pi n$ the left-hand limit is zero while the right-hand limit is $-1$, so $h$ is not continuous at $\pi + 2\pi n$. Finally, at $\pi / 2 + 2\pi n$ we do have that the left-hand and right-hand limit are both $0$, however $h(\pi / 2 + 2\pi n)=1$, so $h$ is not continuous at $\pi / 2 + 2\pi n$.
To see why $h$ is continuous elsewhere, observe that if $x$ lies in an interval of the form $(2\pi n, \pi/2 + 2\pi n)$ or $(\pi/2 + 2\pi n, \pi + 2\pi n)$ then $h(x)=0$, i.e., $h$ is constant, and therefore continuous. Similarly, if $x$ lies in an interval $(\pi + 2\pi n, 2\pi n)$ then $h(x)=-1$, so is continuous. This takes care of all the possibilities, so we are done.
Is this correct? I feel ok about it, but want to make sure I didn't miss something.
Building on my comment, your proof is absolutely fine, but you could make it a bit clearer by saying that $h(x)$ is discontinuous iff $\sin\,x \in \{0, 1\}$ (you have correctly identified the set of $x$ for which that holds). For the second part of the proof, it might be a bit more intuitive and explicit to say that, if $\sin\,x\not\in\{0, 1\}$, then either (a), $0 < \sin\,x < 1$ or (b), $-1 \le \sin\,x < 0$, and, then, because $\sin$ is continuous and never takes on values less than $-1$, we have that for some $\delta > 0$, either (a), $0 < \sin\,y < 1$ or (b), $-1 \le \sin\,y < 0$ for all $y \in (x - \delta, x + \delta)$, so $h$ is constant on a neighbourhood of $x$ ($0$ in case (a) and $-1$ in case (b)). Hence $h$ is continuous at $x$.