Discrete subgroups of $SL(2,\mathbb{C})$ characterization in terms of the norm using "diagonal process".

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I'm reading the book "The Geometry of Discrete Groups" by Alan F. Beardon, and on pages 14 and 15 it gives a characterization of discrete subgroups of $SL(2,\mathbb{C})$ that says:

The subgroup $G$ of $SL(2,\mathbb{C})$ is discrete if and only if for each positive $k$, the set $$\{A\in G : \|A\| \leq k \}$$ is finite.

Here, the norm for the matrix $A\in SL(2,\mathbb{C})$ with coefficients $a,b,c,d\in \mathbb{C}$ is given by $\|A\|=\sqrt{|a|^2+|b|^2+|c|^2+|d|^2}$.

I understand that if the set is finite, then the group $G$ is discrete. The problem is for the converse:

The book says that if the set $\{A\in G : \|A\| \leq k \}$ is infinite, then there are distinct elements $A_n \in G $ with $\|A_n\| \leq k$ for $n\in \mathbb{N}$. If $A_n$ has coefficients $a_n, b_n, c_n, d_n$ then $|a_n| \leq \|A_n \| \leq k$, so the sequence $(a_n)_{n\in \mathbb{N}}$ has a convergent sequence. The same is true for the other sequences of coefficients $(b_n)_{n\in \mathbb{N}},(c_n)_{n\in \mathbb{N}},(d_n)_{n\in \mathbb{N}}$. Until here, everything is clear for me. Now it says:

Using the familiar "diagonal process" we see that there is a subsequence on which each of the coefficients converge. On this subsequence, $A_n \to B$ say, for some $B$ and as $det$ is continuous, $B \in SL(2, \mathbb{C})$, thus $G$ is not discrete.

I don't understand what it means when it says the "diagonal process". I have searched the "diagonal process" and I found that it was the method used by Cantor to prove that the interval $(0,1)$ is uncountable, but I don't understand what this has to do with my problem, or how to apply this method to find such a subsequence.

The rest of the proof is clear too, I just need to understand that part where it says the "diagonal process".

I would be very grateful if someone could explain this to me.

Thank you in advance!

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Diagonal process means that you have sequence $a_n,b_n,c_n,d_n$, extract the a subsequence $a_{n_l}$ from $a_n$ which converges. You can extract a subsequence $b_{n_{l_p}}$ from $b_{n_l}$ which converges, the sequence $a_{n_{l_p}}$ converges also. This process can be continued by extracting a convergent subsequence from $c_{n_{l_p}}$,...