Ultralimit of metric spaces is complete Consider the following short proof that all ultralimits are complete

(Thy typo $d_omega$ is of course to be read as $d_\omega$.)
There are two things that I dont understand about this proof.
1) Why is the sequence $(y_n)$ bounded? Or more precisely why is the sequence of distances $(d_n(y_n,p_n))$ to the basepoints $p_n\in X_n$ bounded?
The argument here looks incomplete to me, because first of all the sequence $(y_n)$ is left undefined on the set $\bigcap N_j$. But this set may be infinite, so the assertion seems to be wrong if we just define it arbitrarily on these points. So I think one should either say that we can pick the sequence $N_j$ such that $\bigcap N_j=\emptyset$ (because $\omega$ is non-princilpe(!)) or that we just assign $y_n=p_n$ for $n\in\bigcap N_j$. However with these fixes it is still not clear to me why the sequence is bounded.
2) Why is $d_\omega(x^I,y)\le2\varepsilon$?
[The proof is from  John Roe. Lectures on Coarse Geometry]
 A: I will rewrite Roe's proof from scratch.   
By induction define a strictly decreasing family of $\omega$-thick subsets $N_j\subset {\mathbb N}$ such that for all $i, i'\in [1, j]$ 
and all $n\in N_j$, we have 
$$
|d_n(x^i_n, x^{i'}_n) -d_\omega(x^i, x^{i'})|< 2^{-j}\le 1. 
$$
Since $\bigcap_{j} N_j=\emptyset$, for each $n\in N_1$ there exists the largest $j=j(n)$ such that $n\in N_j$.  For the same reason, 
$$
\lim_{\omega} j(n)=\infty.  
$$
In particular, for all $n\in N_1$, and all $i, i'\in [1, j(n)]$ we have
$$
|d_n(x^i_n, x^{i'}_n) -d_\omega(x^i, x^{i'})|< 2^{-j(n)}. 
$$
Define a sequence $y_n:= x_n^{j(n)}\in X_n$. The claim is that $(y_n)$ defines a point in $X_{\omega}$ and that 
$$
\lim_{i\to\infty} x^i= y.  
$$ 
I will prove both claims simultaneously. Take $\epsilon>0$. Since $(x^i)$ is a Cauchy sequence in $X_{\omega}$, there 
exists $I$ such that for all $i, i'\ge I$, we have
$$
d_{\omega}(x^i, x^{i'})<\epsilon. 
$$
We can choose $I$ such that, in addition,
$$
2^{-I}<\epsilon. 
$$
I will verify the inequality
$$
d_{\omega}(x^i, y)<2\epsilon, \forall i\ge I.  
$$
Since $x^I\in X_\omega$, it will imply that $y\in X_\omega$; this will also establish the limit  $\lim_{i\to\infty} x^i= y$. 
Fix $i\ge I$. Since $\omega$ is a nonprincipal ultrafilter, and $\lim_{\omega} j(n)=\infty$, 
there exists an $\omega$-thick set $N\subset {\mathbb N}$ such that 
$$
i\le j(n), \forall n\in N. 
$$
By the definition of $d_\omega$ we have: 
$$
d_\omega(x^i, y)=\lim_{\omega} d_n(x_n^i, y_n)= \lim_{\omega} d_n(x_n^i, x_n^{j(n)}). 
$$ 
Hence, it suffices to verify the upper bound 
$$
d_n(x_n^i, x_n^{j(n)})< 2\epsilon,
$$
for all $n$ in some $\omega$-thick subset $M\subset {\mathbb N}$. 
For every $j=j(n)$, $n\in N$, we have $I\le j$. Hence, 
$$
d_\omega(x^i, x^j)<\epsilon 
$$
for such $j=j(n)$, $n\in N$. At the same time, by the definition of the function $j(n)$, 
for all $n\in N_1$, for all  $k, i'\in [1, j(n)]$ we have
$$
|d_n(x_n^k, x_n^{i'}) -d_\omega(x^k, x^{i'})|< 2^{-j(n)}.
$$
I will take $k=i$ and $i':= j(k)$: The inequality $i\le j(n)$ holds for all $n\in N$. Hence, for all $n\in N_1, i\in [1,j(n)]$ 
we have 
$$
|d_n(x_n^i, x_n^{j(n)}) -d_\omega(x^i, x^{j(n)})|< 2^{-j(n)}.
$$
By combining the two inequalities (and using the triangle inequality), for all $n\in M:= N\cap N_1$, we obtain 
$$
d_n(x_n^i, x_n^{j(n)}) < \epsilon + 2^{-j(n)} <2\epsilon,
$$
since $2^{-j(n)}\le 2^{-I} < \epsilon$. qed
