I found alternative proof of Levy's continuity theorem. In spite of reading this proof 12 times and being able to find explanation to therse parts, which were not clear for me, there is 1 crucial point, which I completely fail to understand.
Since this alternative proof is quite long, I provide its beginning to provide context of the part not understood by me.
First of all, let's a recall lemma used in Levy's continuity theorem and Levy's continuity theorem:
Lemma - Let: F - CDF defined on $\mathbb{R}$ with a corresponding to it characteristic function $\varphi$. Then:
$$\forall u \in \mathbb{R}_{+}: F(\frac{2}{u})-F(-\frac{2}{u}) \geq 1 - \int\limits_{-u}^{u}[1-\varphi(s)]ds$$
Theorem(Levy's continuity theorem) - Let: $\{X_{n}\}_{n=1}^{+\infty}$ - sequence of random variables, $\{\varphi_{n}\}_{n=1}^{+\infty}$ - sequence of characteristic functions corresponding to $\{X_{n}\}_{n=1}^{+\infty}$.
If: $\forall$ $t \in \mathbb{R}$: $\lim\limits_{n \to +\infty}[\varphi_{n}(t)]=\varphi(t)$ and $\varphi$ is continous at $t=0$, then: $\varphi$ is a characteristic function of some random variable $X$ and $X_{n} \xrightarrow{\text{d}} X$ (i.e. $X_{n}$ is convergent in distribution in to $X$).
Proof(beginning part): Let: $F_{1},F_{2},F_{3},\dots$ be CDFs corresponding to random variables $X_{1},X_{2},X_{3},\dots$ respectively. The proof will be conducted in few steps.
Step 1: Let: $\epsilon \in \mathbb{R}_{+}$. Due to the continuity of $\varphi(t)$ at $t=0$ and because $\varphi(0)=0$, then at the sufficiently small neighbourhood of zero the values of $\varphi(t)$ differ slightly from 1 and
$$\exists u \in \mathbb{R}_{+}: \frac{1}{u}\int\limits_{-u}^{u}[1-\varphi(s)]ds=\frac{1}{u}\int\limits_{-u}^{u}[1-Re(\varphi(s))]<\frac{\epsilon}{2}$$
By the convergence of $\varphi(t)$ for every $t \in \mathbb{R}$ and by dominated convergence theorem we have:
$$\forall s \in \mathbb{R} \exists n_{0} \in \mathbb{N} \forall n \geq n_{0}: \frac{1}{u}\int\limits_{-u}^{u}[1-\varphi(s)]ds \leq \epsilon $$
By the lemma mentioned at the beginning of this post, we have: $$F_{n}(\frac{2}{u})-F_{n}(-\frac{2}{u}) \geq 1 - \epsilon$$
Step 2: Let: $S_{u}:=\{s_{n} : n \in \mathbb{N}\}$ - the set of all rational numbers in $I_{u}:=[-\frac{2}{u},\frac{2}{u}]$. By the use of the diagonal method choose from $\{F_{n}\}_{n=1}^{+\infty}$ a subsequence $\{F_{n,n}\}_{n=1}^{+/infty}$ convergent to some number, denoted by $F_{u}(s_{i})$ for every $s_{i} \in S_{u}$. $\{F_{n}(s_{i})\}_{n=1}^{+\infty}$ is bounded, hence it contains a subsequence convergent to a point denoted by $F_{u}(s_{1}) \in [0,1]$. Hence, there exists a subsequence $\{F_{1,n}\}_{n=1}^{+\infty}$ of $\{F_{n}\}_{n=1}^{+\infty}$ such that: $$\lim\limits_{n \to +\infty}[F_{1,n}(s_{1})]=F_{u}(s_{1})$$
By considering $\{F_{1,n}(s_{2})\}_{n=1}^{+\infty}$, in analogical way we get a subsequence $\{F_{2,n}\}_{n=1}^{+\infty}$ from $\{ F_{1,n}(s_{1})\}_{n=1}^{+\infty}$ and a number $F_{u}(s_{2}) \in [0,1]$ such that:
$$ \lim\limits_{n \to +\infty}[F_{2,n}(s_{2})]=F_{u}(s_{2})$$
We continue this construction until the exhaustion of $S_{u}$. We obtain a table of CDFs $(F_{k,n})$ $(n,k \in \mathbb{N})$ and a sequence $\{F_{u}(s_{k})\}_{k=1}^{+\infty}$ such that $\{F_{k,n}\}_{n=1}^{+\infty}$ is a subsequence of the sequence $\{F_{m,n}\}_{n=1}^{+\infty}$ if $m \leq k$ and:
$$\lim\limits_{n \to +\infty}[F_{k,n}(s_{m})]=F_{u}(s_{m})$$ for $m \leq k$, $m,k \in \mathbb{N}$
From this we conclude that $\{F_{n,n}\}_{n=1}^{+\infty}$ consists of the elements of main diagonal of the table and satisfies:
$$ \forall s_{k} \in S_{u}: \lim\limits_{n \to +\infty}[F_{n,n}(s_{k})]=F_{u}(s_{k})$$
(...)
What I do not understand is
what is actually meant by $\{F_{n,n}\}_{n=1}^{+\infty}$ in the proof above. In othere words, I can't see how this subsequence is extracted from the sequence $\{F_{n}\}_{n=1}^{+\infty}$, especially the fact how come we suddenly can obtain 2 indices (n,n)...
I would be thankful for providing explanation!!! $\blacksquare$