Here is one definition of totally bounded: A subset $E$ of a metric space $X$ is totally bounded if for every $\epsilon > 0$ there exists a finite subset $\{x_1,...,x_n\}$ of $X$ s.t. $E\subset\cup_{k=1}^nB(x_k,\epsilon)$.
I would like to show that it is equivalent to say that a subset $E$ of a metric space $X$ is totally bounded if for every $\epsilon > 0$ there exists a finite subset $\{y_1,...,y_n\}$ of $E$ s.t. $E\subset\cup_{k=1}^nB(y_k,\epsilon)$: That is, if $E$ is totally bounded as in Def. 1, $E$ is totally bounded as in Def. 2.
For each $x_i$, one could choose some $y_i$ and a ball around $y_i$ with radius at least $\epsilon+\rho(x_i,y_i)$ to get a finite subcover of $E$ consisting of elements of $E$, but of course the radii cannot then be arbitrarily small. I would appreciate a hint for now because I cannot figure out anything more useful than this.
Yes, they are equivalent. Of course, if $E$ is totally bounded in the sense of definition 2, then it is also totally bounded in the sense of definition 1. Suppose now that $E$ is totally in the sense of definition 1. Take $\varepsilon>0$. There is a finite subset $\{x_1,\ldots,x_n\}$ of $X$ such that$$E\subset\bigcup_{k=1}^nB\left(x_k,\frac\varepsilon2\right).$$You can assume without loss of generality that each open ball $B\left(x_k,\frac\varepsilon2\right)$ contains some $y_k\in E$. But then$$B\left(x_k,\frac\varepsilon2\right)\subset B(y_k,\varepsilon)$$and therefore$$E\subset\bigcup_{k=1}^nB\left(y_k,\varepsilon\right).$$