$Y \subseteq(X,d)$
If $Y$ is totally bounded then for each $\epsilon$ there is $\{ y_1,y_2,...,y_n\}$ such that $Y \subset \cup_i^n B(y_i,\epsilon)$.
Now let's say that I want to bound the set Y. I can choose $\epsilon = M $ and there exist un point $y_0 \in Y$ such that $Y \subset$ B$(a,M)$ because $Y$ is totally bounded.
Is this right?
The claim is true, but your argument is not valid. If you take, say, $\varepsilon =1$, then the definition of totally bounded does not give you a single $a\in X$ with $Y\subseteq \mathrm B(a,1)$. But, you are very close. Work correctly with the definition, and you'll get a bunch of finitely many balls of radius $1$ which together cover the entire space. Do something with them to get your result.