I'm given $f:[0,1]\rightarrow \mathbb{R}$, $f$ continuous, takes a non-negative value somewhere in $[0,1)$ and is strictly negative at $1$, i.e. $f(1)<0$. By IVT we know the solution set of $f(x)=0$ is non-empty. My question is, can we show it is compact? I saw examples such as this which use differentiability (which I don't have) but also requires finiteness of the solution set (which I don't need).
Any help is appreciated.
Yes. We see that $\{x \in [0, 1] | f(x) = 0\}$ is bounded. It is also closed, since it can be written as $f^{-1}(\{0\})$ and $\{0\}$ is closed. A closed and bounded subset of $\mathbb{R}$ is compact.
Conversely, any compact set $C \subseteq [0, 1]$ is a solution set to $f(x) = 0$ for some continuous $f$. In this case, define $f(x) = \min\limits_{c \in C} |c - x|$ if $C$ is non-empty, $f(x) = 1$ otherwise. Then $f$ is continuous and $f(x) = 0$ iff $x \in C$.