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]
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