Determine why these sets are necessarily open, closed or compact where $f$ is a continuous function

Question

Determine why these sets are necessarily open, closed or compact where $f$ is a continuous function

$f : \mathbb{R} \to \mathbb{R}$

a) $S=$ { $x \in \mathbb{R} : f(x) = -3$ }

{$-3$} is closed thus $S = f^{-1}(\{-3\})$ is closed, the singleton {$ - 3$} is compact but I don't know if that gives that $S$ is compact

b) $S = \{ x \in \mathbb{R} : f(x) > 1\}$, $f(S) = (1, \infty)$ is open so $S$ is open

c) $S = \{ x \in \mathbb{R} : 0 \le f(x) <1\}$ This one seems tricky since $f(S) = [0,1)$ which is not open and not closed so it doesn't seem to be any of the choices.

Answer 1

To break compactness, consider constant functions. To break open/closedness in (C), you can consider the identity function!