This is theorem 7.23 in Baby Rudin.
If $\{ f_n \}$ is pointwise bounded sequence of complex functions on a countable set $E$, then $\{ f_n \}$ has a subsequence $\{ f_{n_k} \}$ such that $\{ f_{n_k}(x) \}$ convverges for every $x \in E$ (i.e. it has a pointwise convergent subsequence).
The proof says that $\{ f_{n}(x_1 \}$ is bounded so there exists a subsequence $\{ f_{1,k} \}$ such that $\{ f_{1,k}(x_1 \}$ converges as $k \to \infty$. I understand that. And I also understand the next part which says that to repeat this process and create a "matrix" with rows of these sequences. But then comes the part that I don't understand. Rudin then writes that we need to go down the diagonal of this "matrix" and choose $\{ f_{1,1}, f_{2,2}, f_{3,3}, \dots \}$ because this sequence will be our pointwise convergent subsequence. Why do we need to go along the diagonal and why can't we, for example, take the first "column", i.e. $\{ f_{1,1}, f_{2,1}, f_{3,1}, \dots \}$? It feels like the first column would still fulfill all the conditions that he sets, but obviosuly I am wrong but I can't see why