Let's say we have a set $S$. Is there a sequence $u:\mathbb N \to S$ such that every other sequence $v$ is a subsequence of $u$?
Here is what I have so far:
If $S=\emptyset$, then no (there are no sequences to begin with).
If $|S|=1$, then yes (there is only one sequence, and every sequence is a subsequence of itself).
After these two trivial cases, I got stuck. (I expect any of these universal sequences to be very interesting though.) I know that the relation of subsequence is a preorder.
For similar constructs, see Rado graph or surreal numbers.
For $\mathbb N$, consider the sequence $0, 0, 1, 0, 1, 2, 0, 1, 2, 3, 0,1,2, 3, 4, \dots$.
I claim this sequence contains every $\mathbb N$-valued sequence as a subsequence.
This solves all countable cases. The uncountable cases are clearly impossible, since your sequence will not even cover every element, let alone every subsequence. (But if you allow uncountably long sequences, a similar trick will work.)