The composite of all unramified extensions inside an algebraic closure

Question

I'm reading Ch.II, $\S$ 7 of Neukirch's Algebraic Number Theory and I'd be really grateful if someone could help me understand the following:

Let $K$ be a complete valued field wrt a non-archimedean valuation $v$. Neukirch proves that the composite of two finite unramified extensions of $K$ is again unramified.

How does it follow from this that the composite of all unramified extensions of $K$ inside a fixed algebraic closure $\bar{K}$ is again unramified (as he is assuming on p. 154)?

The problem for me is that there could be an infinite number of such subextensions to consider.

Many thanks for your answers.

Answer 1

I think that he is trying to employ the lemma of Zorn. And to show that the composite of two unramified extensions is still unramified amounts to verifying the condition required for that lemma. The result follows directly from the lemma.
Maybe this should be put into the comment-form? Thanks in any case.

Answer 2

By definition (see 7.1 in p. 153 of Neukirch's book), an arbitrary algebraic extension $L/K$ is unramified if it is the union of finite unramified subextensions. So the compositum of all finite unramified extensions of $K$ is, by definition, unramified.