I try to prove that every subset of a totally bounded set is totally bounded, but I am not really sure with my argument here. Could someone help me check it?

Suppose A is a subset of a totally bounded set X. Since X is totally bounded, there exists an $\displaystyle \frac{\epsilon}{2}$-net $\displaystyle F=\{x_1,x_2,..,x_n\}$ that covers X, i.e $\displaystyle X\subset \bigcup \{B(x_i,\epsilon/2):x_i\in F\}$. Since A is a subset of X, A is also a subset of the union above. So, $\displaystyle A\cap B(x_i,\epsilon/2)$ for some $\displaystyle x_i \in F$. Let $\displaystyle y\in A\cap B(x_i,\epsilon/2)$. Then $\displaystyle B(y,\epsilon)$ covers A since this ball contains every ball with radius $\displaystyle \epsilon/2$ above. Something does not sound right here since I am not sure if the new ball with radius $\displaystyle \epsilon$ contains all the balls with radius $\displaystyle \epsilon/2$