Subset of totally bounded set is totally bounded.

So $B\subseteq\bigcup_{i=1}^{m}B_{\epsilon/2}(a_{i})$ by choosing $\epsilon/2$-cover for $A$, and assume without loss of generality that $B_{\epsilon/2}(a_{i})\cap B\ne\emptyset$ for all $i=1,...,m$. Pick a $b_{i}\in B_{\epsilon/2}(a_{i})\cap B$. We claim that $\{B_{\epsilon}(b_{i})\}_{i=1}^{m}$ cover $B$:

For $b\in B$, pick an $i$ such that $b\in B_{\epsilon/2}(a_{i})$. Then $d(b,b_{i})\leq d(b,a_{i})+d(a_{i},b_{i})<\epsilon$, so $b\in B_{\epsilon}(b_{i})$, as expected.

Suppose $B \subseteq A$ and let $\varepsilon > 0$. Find a finite covering of $A$ by balls of radius $\frac{\varepsilon}{2}$, say $$B_{\varepsilon/2}(a_1),\ \dots,\ B_{\varepsilon/2}(a_n)$$ where $a_1,\dots,a_n \in A$.

By relabelling the elements $a_1,\dots,a_n$ if necessary, you can assume that the first $k$ of these balls contains an element of $B$, and the remaining $n-k$ do not. For each $1 \le i \le k$, choose some $b_i \in B_{\varepsilon/2}(a_i)$, and note that $B_{\varepsilon}(b_1),\dots,B_{\varepsilon}(b_k)$ is a cover of $B$.