Diagonalization argument for convergence in distribution Let $Z_{n, N}$ be a sequence of random variable such that, for any fixed $N$, this sequence of variables converges in distribution to a variable $Z$ as $n \rightarrow \infty$. There exists a sequence $N(n)$ that goes to infinity ($N(n) \rightarrow \infty$ as $n \rightarrow \infty$) such that $Z_{n, N(n)}$ converges to $Z$ in distribution as $n \rightarrow \infty$? How to construct the sequence $N(n)$ and prove the convergence in distribution of the resulting random variables $Z_{n, N(n)}$?

A little bit of context: An argument similar to the one above is used in Terence Tao, "Topics in Random Matrix Theory" book under the name of "diagonalization argument". In Section 2.2.1, the argument is used to show the possibility of considering bounded random variables to prove the central limit theorem without loss of generality. There $N$ is the upper bound for the variables and it is assumed that since central limit theorem yield $Z_{n, N} \rightarrow \mathcal{N}(0, 1)$, for any fixed $N$, it hold that $Z_{n, N(n)} \rightarrow \mathcal{N}(0, 1)$ for some sequence of bounds $N(n)$ dependent on $n$.
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 A: Here is another argument that combines Antonio's approach of dense sets with some of my comments:
Given:
Let $F(x)$ be a CDF. Let $S$ be the (at most countably infinite) set of discontinuities of $F$.  Let $F_{n,k}$ be a collection of CDFs indexed by $(n,k) \in \mathbb{N}^2$ such that
$$ \lim_{n\rightarrow\infty} F_{n,k}(x) = F(x) \quad \forall x \in \mathbb{R} \setminus S, \quad \forall k \in \mathbb{N}$$
Construction:
Let $D$ be the set of rational numbers in $\mathbb{R} \setminus S$.  Let $\{x_1, x_2, x_3, \ldots\}$ be an ordering of $D$.
For each $i$ and each $k$ there is some positive integer $G(i,k)$ such that
$$|F_{n, i}(x_k) - F(x_k)|\leq 1/i  \quad \forall n \geq G(i,k)$$
Define $G(1)=1$.  For $i \in \{2, 3, 4, ...\}$ define
$$ G(i) = \max\{G(i,1), G(i,2), ..., G(i,i)\}$$
Then for all $i \geq 2$ we get:
$$ |F_{n,i}(x_k)-F(x_k)|\leq 1/i \quad \forall n\geq G(i), \forall k \in \{1, \ldots, i\}$$
For each positive integer $n$ define $N(n)$ as the largest $i \in \{1, ..., n\}$ such that $G(i)\leq n$ (such a value $i \in \{1, ..., n\}$ always exists because $G(1)=1\leq n$). It follows that whenever $N(n)\geq 2$ we get
$$ \boxed{|F_{n,N(n)}(x_k)-F(x_k)|\leq 1/N(n) \quad \forall k \in \{1, \ldots, N(n)\}}$$
Claim 1: $N(n)\rightarrow \infty$.
Proof: Fix $j\in \{1, 2, 3, ...\}$.  Then if $n$ is any positive integer that satisfies
$$ n \geq \max[j, G(j)]$$
we have that
$$j \in \{i \in \{1, ..., n\}  : G(i)\leq n\}$$
Since $N(n)$ is defined as the largest value in the set $\{i \in \{1, ..., n\}  : G(i)\leq n\}$, we get  $N(n)\geq j$. $\Box$
Now fix $k \in \{1, 2, 3, ...\}$. For sufficiently large $n$ we get $N(n)\geq 2$ and $k\leq N(n)$. So taking a limit of the boxed equation as $n\rightarrow\infty$ gives
$$\lim_{n\rightarrow\infty} |F_{n,N(n)}(x_k)-F(x_k)| = 0 $$
This holds for all $x_k \in D$, from which we should also get convergence for all $x \in \mathbb{R} \setminus S$.
A: I managed to think of a tentative argument after Michael's comment. So I post here in case someone falls into the same question as me...

As Michael suggested, it is easier to start by proving the result for sequences of real numbers. If $z_{n, N}\in \mathbb{R}$ is a double sequence such that $z_{n, N} \rightarrow z$ for every $N$. Defined the following sequence
\begin{equation}
N(n) = \inf \left\{i:  |z_{\ell,i} - z| > \frac{1}{i} \text{ for all } \ell \ge n\right\}
\end{equation}
Except for the trivial case: $z_{\ell,i} = z$ for $\ell>n$ for all $i$; the set above can be empty (and $N(m) = \infty$) only for finitely many values. We can just arbitrarily replace these values without affecting the convergence.
Such a sequence goes to infinity. For an arbitrary $i$, we have that $|z_{\ell,i} - z| \le \frac{1}{i}$ for all $l \le n$, hence $N(n)>i$, for sufficiently large $n$. Moreover, $z_{n, N(n)} \rightarrow z$ as $n \rightarrow \infty$.

Now let $F_{n, N}(x)$ be the cumulative distribution function (CDF) of the random variable $Z_{n, N}$. And let $F(x)$ be the CDF of the random variable $Z$. For every fixed $N$, since $Z_{n, N}$ converges in distribution to $z$, $F_{n, N}(x)\rightarrow F(x)$ at every $x\in \mathbb{R}$ at which $F$ is continuous.
Now, using the other direction of the equivalence, to show convergence in distribution of $Z_{n, N(n)}$ to $Z$ it is sufficient to show that $F_{n, N(n)}(x)\rightarrow F(x)$ for every $x\in \mathbb{R}$ at which $F$ is continuous.
Since $F$ is a monotonic function it has at most countably many discontinuities, say, $\mathcal{S}$.
Now pick a countable and dense subset of $\mathbb{R} - \mathcal{S}$,  and call the points in this set of $x_m$. Define,
\begin{equation}
N(n) = \inf \left\{i:  |F_{\ell,i}(x_m) - F(x_m)| > \frac{1}{i} \text{ for all } 1 \le m \le 1 \text{ and }\ell \ge n \right\}
\end{equation}
using the same argument as before we have that $F_{n, N(n)}(x_m)\rightarrow F(x_m)$ for every $m$. The convergence in a dense subset of $\mathbb{R}-\mathcal{S}$ implies the convergence in all continuity points of $F$.

Obs: $\mathbb{R} - \mathcal{S}$ is separable (has countable dense subset) because metric spaces are hereditarily separable (see If $X$ is a separable metric space and $M \subset X$ is a metric subspace then $M$ is separable.)
