Let $f: \mathbb{R}^{n} \to \mathbb{R}$ be a continuous function and $n \geq 2$. Moreover, suppose that there is $c \in \mathbb{R}$ such that $f^{-1}(\{c\}) = \{x \in \mathbb{R}^{n} \mid f(x) = c\}$ is bounded.
(a) Prove that $\{x \in \mathbb{R}^{n} \mid f(x) \geq c \}$ or $\{x \in \mathbb{R}^{n} \mid f(x) \leq c \}$ is bounded.
(b) Prove that $f$ has a maximum or minimum in $\mathbb{R}^{n}$.
My attempt.
(a)Suppose that both are unbounded. Just take $r > 0$ such that $B_{r}(0) \supset f^{-1}(\{c\})$. Since $n \geq 2$, $\mathbb{R}^{n}-B_{r}(0)$ is connected. But $f$ is continuous, then apply the IVT for conclude that $f^{-1}(\{c\})$ is unbounded, a contradiction.
(b) Suppose, WLOG, that $F = \{x \in \mathbb{R}^{n} \mid f(x) \geq c \}$ is bounded. So, $F$ has a maximum or minimum, namely $f(p)$. If $f(p)$ is a maximum, its easy. If $f(p)$ is a minimum, then we should be $f(p) = c$, since $f(\mathbb{R}^{n})$ is connected. So...?
I dont know how to finish the proof. Can someone help me?
Because $F$ is bounded, it is compact, so $f$ attains a maximum on $F$, which is a global maximum of $f$.