I need to prove the following:
Let $f:\mathbb{R}^n\rightarrow \mathbb{R}$ be continuous on all of $\mathbb{R}^n$.
$f$ is coercive $\iff \forall \alpha\in\mathbb{R}.\left\{x \mid f(x) \leq \alpha \right\}$ is compact.
you can assume for the definition of coercive:
$$\lim_{\|x\| \to \infty} f(x) = \infty.$$
This is my proof:
Since $f$ is continuous this implies the closedness of the sets $\{x|f(x)\leq\alpha\}$. It remains to show that any set of the form $\{x|f(x)\leq\alpha\}$ is bounded.
Suppose that there is an $\alpha\in \mathbb{R}^n$ such that the set $S = \{x|f(x)\leq\alpha\}$ is unbounded. There there must exist a sequence $\{x_n\}\subset S$ with $||x_n||\rightarrow\infty$. But by coercivity of $f$ we must also have that $f(x_n)\rightarrow\infty$. This contradicts the fact that $f(x_n)\leq\alpha$ for all $n = 1,2\ldots$. Therefore $S$ must be bounded.
Assume that each of the sets $\{x|f(x)\leq\alpha\}$ is bounded and let $\{x_n\}\subset\mathbb{R}^n$ be such that $||x_n||\rightarrow\infty$. Suppose that there exists a subsequence of the integers $J\subset\mathbb{N}$ such that the set $\{f(x_n)\}$ is bounded above. There there exists an $\alpha\in\mathbb{R}^n$ such that $\{x_n\}_J\subset \{x|f(x)\leq\alpha\}$. This cannot be the case since each of these sets is bounded while every subsequence of$\{x_n\}$ is unbounded by definition. Therefore, the set $\{f(x_n)\}_J$ cannot be bounded, and so $\{f(x_n)\}$ contains no bounded subsequence i.e $f(x_n)\rightarrow\infty$.