$G\colon\mathbb{R}\to\mathbb{R}$ is a continuous function. Suppose that $\exists m\in\mathbb{R}$ such that $\{x\in\mathbb{R}: G(x)\leq m\}$ is bounded and not empty. Show that $G$ has an absolute minimum.
I'm not sure how to approach this exercise. I believe that Weierstrass' Theorem may play a part somehow, but I don't know how. I so am not sure what to do about the bounded and not empty set, or how it plays a role.
Let $M$ be the set of those real numbers $x$ such that $G(x)\leqslant m$. It is non-empty and bounded. Let $a$ and $b$ be its infimum and supremum respectively. Since $G$ is continuous, its restriction to $[a,b]$ attains an absolute minimum at some point $x_0$. And outside $[a,b]$ you always have $G(x)>m$. Therefore, $G$ has an absolute minimum at $x_0$.