Is there $\sigma_\infty$ such that $\sigma_\infty$ is a subsequence of $\sigma_n$ for all $n \in \mathbb N$? I have come across this "Diagonal method" in this lecture note.

Theorem: Let $S$ be a non-empty set. For each $n\in \mathbb N$, let $\sigma_n$ be a sequence of elements of $S$. We denote by $\sigma_{n}(m)$ the $m$-th coordinate of $\sigma_n$. Assume that $\sigma_{n+1}$ is a subsequence of $\sigma_n$ for all $n\in \mathbb N$. Then there is a sequence $\sigma_\infty$ such that the restriction
$$
\sigma_\infty \restriction \{n, n+1, \ldots\}
$$
is a subsequence of $\sigma_n$.

Proof: Clearly, the sequence $\sigma_\infty$ with $\sigma_\infty (n) := \sigma_n(n)$ satisfies the requirement.
I think a stronger statement i.e.,

there is a sequence $\sigma_\infty$ such that $\sigma_\infty$ is a subsequence of $\sigma_n$ for all $n \in \mathbb N$,

could not hold in general. Could you provide me with a counter-example of above statement?
 A: Let $\sigma_n(m) = n+m$ (so $S = \mathbb N$, for example).

Claim : $\sigma_{i+1}$ is a subsequence of $\sigma_i$ for all $i 
\geq 1$. Furthermore, there is no sequence $\sigma_{\infty}$ such that $\sigma_{\infty}$ is a subsequence of $\sigma_i$ for all $i \geq 1$.

Proof : One checks that $$\sigma_{i+1}(k) = (i+1)+k = i+(k+1) = \sigma_i(k+1)$$
for all $i , k \geq 1$. Therefore, $\sigma_{i+1}$ is a subsequence of $\sigma_i$, in particular if $f(i) = i+1$ then $\sigma_{i+1}(\cdot) = \sigma_{i}(f(\cdot))$.
Finally, if $\sigma_{\infty}$ is a sequence which is a subsequence of every $\sigma_i$, then consider the element $\sigma_{\infty}(1)$. It certainly belongs to $\mathbb N$, otherwise the subsequence property can't hold since the image of every $\sigma_i$ lies in $\mathbb N$. If $\sigma_{\infty}(1) = K \in \mathbb N$, then obviously it cannot be a subsequence of $\sigma_{K+1}$, whose image doesn't contain $K$. We are done.

You may have wondered why I took the natural numbers above and , perhaps, why not a simpler set (At least I wondered about this!). The answer to that is : if $S$ is finite then the desired property is in fact , true. One can show that using a fairly clever trick. More precisely :

If $S$ is finite and $\sigma_i$ is a sequence of sequences with values in $S$ with $\sigma_{i+1}$ being a subsequence of $\sigma_i$ for all $i$, then there exists $\sigma_{\infty}$ , a sequence which is a subsequence of all the $\sigma_i$.

Let's introduce a notation : Let $[a,b]$ denote the set of all natural numbers between $a$ and $b$, with both included. For example, $[3,7]= \{3,4,5,6,7\}$.
Proof : For each $s \in S$, let $$I_s = \{i \in \mathbb N :|\{k : \sigma_i(k) = s\}| = \infty\}$$ In other words, the set of all $i$ such that $s$ occurs infinitely many times in $\sigma_i$.
Since $\sigma_{i+1}$ is a subsequence of $\sigma_i$ for each $i$, it follows that if $j+1 \in I_s$ then $j \in I_s$ for all $j \in \mathbb N, s \in S$. Therefore, every $I_s$ is either of the form $[1,J]$ for some $J$, or $I_s = \mathbb N$.
Now, suppose that $I_s = \mathbb N$ for some $s \in S$. By definition, the sequence $\sigma_{\infty} \equiv s$ (constant sequence) is a subsequence of all the $\sigma_i$.
We claim that for some $s$ it must happen that $I_s = \mathbb N$. Suppose not : then, every $I_s = [1,J_s]$ for some $J_s$. Let $J = \max\{J_s : s \in S\}$, which is finite because $S$ is finite.
Look at $\sigma_{J+1}$ : this is a sequence in which none of the elements of $S$ occur infinitely many times. That's clearly impossible : $\sigma_{J+1}$ is a map from $\mathbb N$ to $S$ i.e. an infinite set to a finite set, therefore at least one element must have infinite preimage.
Therefore, $I_s = \mathbb N$ for some $s \in S$, and $\sigma_{\infty}$ as described before does the job.$\blacksquare$
