Let (X,p) be a metric space and let S Í X be totally bounded. Argue that the closure of S is totally bounded.
Suppose that $\displaystyle \overline{S}$ is not totally bounded.
There is $\displaystyle c>0$ such that there is no $\displaystyle c\text{-net}$ for $\displaystyle \overline{S}$.
But there must a $\displaystyle c\text{-net}$ for $\displaystyle {S}$.
So some $\displaystyle y\in\overline{S}$ such that $\displaystyle y$ is not in $\displaystyle \bigcup\limits_{k = 1}^n {B(\alpha _k ;c)} $ the $\displaystyle c\text{-net}$ for $\displaystyle S$.
Let $\displaystyle d=\min\{d(y,\alpha_k)-c\},~k=1,2,\cdots n$.
You need to argue that $\displaystyle d>0$.
But $\displaystyle B(y;d)$ must contain a point in $\displaystyle S$.
Now prove that is a contradiction.