No. Take S discrete (meaning: x < y is never true). Then the only chains are singletons.

You will object that this is a fake example and you'd like any two elements to be comparable to something else (or maybe by a finite sequence of such inequalities). Fine.

Take the set of infinite sequences of 0's and 1's. Say a sequence is "finite" if it is eventually zero, so (1,1,1,1,0,0,0,...) is finite but (0,0,0,0,1,1,1,...) is not. If a sequence is finite, say its "initial string" is the sequence up through the last one. In the first case the initial string is (1,1,1,1) and for (1,0,1,0,1,0,0,0,...) the initial string is (1,0,1,0,1).

Say x < y iff x is finite and y begins with the initial string of x. So (1,0,1,0,1,0,0,0,...) < (1,0,1,0,1,1,1,0,1,....)

There are uncountably many different countable chains, and (0,0,0,...) is the minimum of this poset. But the set of infinite sequences is uncountable by Cantor's argument.