Diagonalization method by Cantor (2) I asked a while ago a similar question about this topic. But doing some exercises and using this stuff, I still get stuck. So I have a new question about this topic. (Here is the link for the previous question: Cantor diagonalization method for subsequences).
Now my new question: Suppose I have a sequence $(x_n)$, a set $K$ and a function $f$ and we define for all $l \in \mathbb{N}$ the set $M_l:= K\cap \{x|f(x)\le l\}$ with $f< \infty$ on $K$.
I've proved that for a fixed $l$ there's a subsequence $(x_{n_k})$ converging on $M_l$, denote this by $(x^l_{n_k})$.
First it's clear that $M_l \subset M_{l+1}$ and I want to show that there's a subsequence which converges on $\cup_{l\ge1} M_l = K$.
So I have different subsequences $(x^1_{n_k}),\dots,(x^p_{n_k}),\dots$ and define the diagonal sequence as ${x^{\phi(l)}_{n_{\phi(l)}}}$ where $\phi(l)$ is the $l$-th element of $n_k$ (just the diagonal sequence).
Is it enough to say, since $(M_l)$ is increasing the sequence converges on $K=\cup_{l\ge1} M_l$?
I'm sorry, but I'm just confues about picking the right sequence and to prove that it is the right sequence using Cantor. Thank you for your help

Denote by $(x_{\varphi_l(k)})$ a subsequence which works for $M_l$. In fact, you have to construct these subsequence by induction, in order to make $(x_{\varphi_{l+1}(k)})$ a subsequence of $(x_{\varphi_l(k)})$. Then we put $x_{n_k}=x_{\varphi_k(k)}$. Now we are sure that the sequence $(x_{n_k})_{k\geq N(j)}$ is a subsequence of $(x_{\varphi_j(k)})_{kgeq N(j)}$ for some integer $N(j)$. 
It's important that the subsequences are nested, otherwise it may not work. For example, we assume that for $l$ even only a subsequence of the form $(x_{2k})$ work and for $l$ odd a subsequence of the form $(x_{2k+1})$. Then $x_{n_{\varphi(2l)}}^{\phi(2l)}=x_{4l}$ and $x_{n_{\varphi(2l+1)}}^{\phi(2l+1)}=x_{2l(2l+1)+1}$ so the sequence of indexes is not increasing.
