Let $Y$ be totally bounded subset of metric space $(X,d)$. Then every sequence $(y_n)\in Y$ has a Cauchy subsequence.
Here I have understood the proof until the last step.
How are we getting the value of/upper bound of : $d(y_n^{(n)}, y_{n+1}^{(n+1)})$ ? and same thing for the subsequent terms in the same line aswell.
You estimate this distance by construction: $Y$ being totally bounded means that, for every $\epsilon > 0$ there exists a finite collection of open balls of $Y$ with diameter $\leq 2\epsilon$. If you get, for instance, $\epsilon = \frac{1}{2}$, there exists $a_{1}, ..., a_{d}$ such that $Y \subset \displaystyle\bigcup_{j=1}^{d}B(a_{j}, \epsilon)$. As there are infinite different terms in the sequence $(y_{n})$ there exists one of these balls with infinite elements of $(y_{n})$. Relabel these elements by $(y_{n}^{1})$, then \begin{eqnarray} d(y_{n}^{1}, y_{m}^{1}) \leq d(y_{n^{1}}, a_{j}) + d(y_{m}^{1}, a_{j}) < \frac{1}{2} + \frac{1}{2} \end{eqnarray} for some $a_{j}$ and for every term of this new sequence. Now you repeat this construction with $\epsilon = \frac{1}{2^{2}}, \frac{1}{2^{3}}, ...$ and taking subsequences of $(y_{n}^{1}), (y_{n}^{2}), ...$ and you'll get the result you want. (Edit: When you get the subsequence $(y_{n}^{2})$ this sequence is, in particular, in the ball of diameter $2 \times \frac{1}{2}$, this is the point you are missing to understand the prove)