Originally Posted by

**Drexel28** What's to stop you from doing this:

Let $\displaystyle M=d(\bold{0},F)$ and let $\displaystyle K=B_{M+1}[0]\cap F$ (where $\displaystyle B_{M+1}(0)$ is the closed ball). Then, $\displaystyle K$ is closed since it is the intersection of two closed sets and it is bounded since $\displaystyle \text{diam }K\leqslant 2(M+1)$, thus compact. Also, it is not hard to prove that $\displaystyle \inf_{x\in F}\|x\|\geqslant \inf_{x\in K}\|x\|$. But, $\displaystyle \|\cdot\\:\mathbb{R}^n\to\mathbb{R}:\bold{x}\mapst o\|\bold{x}\|$ is continuous and since $\displaystyle K$ is compact we have that $\displaystyle \|\cdot\|:K\to\mathbb{R}$ assumes a minimum on $\displaystyle K$. The conclusion follows.