Understanding a proof that bounded sequences in $\mathbb{R}^p$ has a convergent subsequence I'm having trouble concerning the following proof that each bounded sequence in $\mathbb{R}^p$ has a convergent subsequence. We have already established that this is true in $\mathbb{R}$ and this is the proof that the professor gives:

Use induction on $p$. 
For $x \in \mathbb{R}^p$ write $x = (y, z)$ with $y = (x^{(1)}, \dots, x^{(p-1)}) \in \mathbb{R}^{p-1}$ and $z = x^{(p)} \in \mathbb{R}$.
Observe that $|x|^2 = |y|^2 + z^2$. $(i)$
Let $x_n = (y_n, z_n)$ for $n \in \mathbb{N}$ be a bounded sequence in $\mathbb{R}$ (I think this is a typo and should be $\mathbb{R}^p$?). From $(i)$, $y_n$ and $z_n$ are also bounded.
Replacing $x, y, z$ with $x_n - x, y_n, - y, z_n - z$ in $(i)$ yields $x_n \to x <=> y_n \to y $ and $ z_n \to z$
By the base case $p = 1$ (established already), some subsequence $z_{n_k} \to z$ for some $z \in \mathbb{R}$, and by the inductive step some subsequence of $y_n$ converges to some $y \in \mathbb{R}^{p-1}$. The indices of the two subsequences are not necessarily the same, so we can't yet conclude that $x_{n_k} \to x = (y, z)$. However, since $y_{n_k}$ is bounded, it itself has a convergent subsequence (subsubsequence). So $y_{n_{k_l}} \to y$ for some $y \in \mathbb{R}^{p-1}$. As a subsubsequence of a subsequence converging to $z$, we also have $z_{n_{k_l}} \to z$
Thus, $x_{n_{k_l}} \to x = (y, z)$. $\blacksquare$

The following are my confusing about the proof:


*

*Why is the substitution $x, y, z \to x_n-x, y_n - y, z_n-z$ justified?

*How does finding a subsequence of the subsequence of $y_n$ and $z_n$ fix the problem of the indices being different?
 A: Yes, the boldface part is probably a typo.
Lets look at 1. first. To prove that a sequence $\{a_n\}$ in $\mathbb{R}^p$ converges to $a$, it is equivalent to show that the sequence $\{a_n-a\}$ converges to $0$. This can be done coordinatewise: $a_n = (a_n^{(1)},\ldots,a_n^{(p)})$ converges to $(a^{(1)},\ldots,a^{(p)})$ if and only if $a_i^{(k)}-a^{(k)}\to 0$ for each $k$ as $i\to\infty$. It's not really a big deal here, since you still go about finding a convergent subsequence without using this fact.
Now on to 2. Your professor may have been a bit messy here, or maybe trying to highlight a part of the proof where the naive thing doesn't work. If one tries the naive thing, then at this step of the proof one has a convergent subsequence $z_{n_k}$ of $z_n$ from the base case, and since $y_n$ is bounded by the induction hypothesis $y_n$ has a convergent subsequence, call it $y_{m_j}$, and these indices have nothing to do with one another, so it's impossible to say anything about the convergence of $x_n$ using the indices $m_j$ or $n_k$. (In fact, if $m_j$ is just the even numbers and $n_k$ is the odd numbers, then they really have nothing to do with one another!)
On the other hand, ignore $z_n$ for now. From the inductive hypothesis there is a convergent subsequence $y_{m_j}$ of $y_n$. Then using this index, $z_{m_j}$ is a subsequence of $z_n$. Since $z_{m_j}$ is bounded, it has a convergent subsequence $z_{n_k}$. Since $y_{n_k}$ is a subsequence of $y_{m_j}$ and $y_{m_j}$ is convergent, $y_{n_k}$ is a convergent subsequence of $y_n$. One has to do it this way to avoid the indices being different. (One could also do this argument with the $z$-s and $y$-s switched.)
