I think I have the idea in my head, but I don't know if I am expressing it clearly in math terms.

The closure of $\displaystyle S$, $\displaystyle \bar{S}$ is defined to be $\displaystyle S \cup S'$ (where $\displaystyle S'$ are the limit points of $\displaystyle S$)

$\displaystyle S$ totally bounded implies that $\displaystyle \forall \epsilon > 0 \; \exists$ finite number of points $\displaystyle x_1, ... , x_n \in S$ s.t

$\displaystyle S = \cup_{k=1}^n \; B(x_k,\epsilon)$

So we already have all the points in S, i.e the union of the balls is a cover for S. We just need to wory about S', the limit points.

$\displaystyle x \in S' \rightarrow \forall \epsilon > 0 \;, \exists y \in S$ s.t $\displaystyle d(x,y) < \epsilon$

so if any limit point is less than $\displaystyle \epsilon$ away from a point in S. We can say

$\displaystyle \bar{S} = \cup_{k=1}^n \; B(x_k,2\epsilon)$

Right? Is my thinking correct? The limit points are close to the set so we can make sure we get them by making the radius bigger.