This should be a pretty known proof in analysis for many of you but I've (I think) proved it in a different way than I've seen in any textbook.
The bounded proof is pretty straightforward and I proved it in a similar way to what Abbott does on Understanding Analysis. Let $\mathcal{C}=\cup_{x\in F}\, I_x$ be an open cover of $F\in\mathbb{R}$, where $I_x = V_1(x)$ for $x\in F$ ($V_1(x)$ means open ball of radius $1$ with center $x$). By hypothesis, there exists a finite subcover $\mathcal{C}_1 \subset \mathcal{C}$ of $F$. $\mathcal{C_1}$ is bounded, so $F$ has to be bounded as well.
Now for the closed part, which is what's different from what I've seen. Suppose instead that $F$ is not closed and let's prove that there exists an open cover of $F$ that does not admit a finite subcover. Let $\mathcal{C}=\cup_{k=1}^\infty\, I_k$, where $I_k = (\inf F + \frac1k, \sup F – \frac1k)$. This is an open cover of $F$ that does not admit a finite subcover, so we should be done here… perhaps?
Best Answer
Let $\overline{F}$ be the closure of $F$. If $F$ is not closed, then $\exists x \in \overline{F} \setminus F$. Define $U_n= \{ y~|~d(x, y) \gt \frac 1n \} $. Then $x \notin F \Rightarrow F \subseteq \bigcup_n U_n$ and $x \in \overline{F} \Rightarrow \forall n \exists y \in F \text{ such that } d(x, y) \lt \frac 1n $, so $ \{ U_n~|~n \in \Bbb N \} $ is an open cover of $F$ with no finite subcover. Notice that this proof works in any metric space with the metric topology.
I think that's the concept you're trying to get at, but you don't know that $\inf(f) \notin F$.