I came across this statement and it is very intuitive to me. I am looking for more information such as a name associated with this theorem or even a more general theorem.
Does such a name or more general theorem exist?
I will sketch a proof of the statement if anyone is interested:
For any element $a$ of the chain there are either an infinite number of elements greater than $a$ or an infinite number of elements less than $a$. If not, the chain is finite.
WLOG, assume that there are an infinite number of elements greater than $a$. Let $f(a) = 1$. Find one of the elements greater than $a$ (call it $a_2$) and if there are an infinite number of elements greater than it, write $f(a_2) = 2$. Continue the construction indefinitely (and show that it is possible to do so). Additionally, it must be shown that the isomorphism is bijective.
I don't quite know the name of the theorem. But I am familiar with a more general principle known as the cofinality principle, stating that in every linear order there is a cofinal well-ordering ($A$ is cofinal in $B$ if for every $b\in B$ there is some $a\in A$ such that $b\leq a$).
Now suppose that $A$ is an infinite linear order. If $A$ has a cofinality of $1$ (namely there is a maximal element), we can remove the maximal element, then by recursion reapply the theorem, and either obtain an infinite decreasing chain, or reach to a linear order without a maximal element.
So suppose now that $A$ has no maximal element, we can pick some element in $a$, then choose the next one and so on. Since there is no maximal element we can always continue choosing elements. And therefore this defines an embedding from $\Bbb N$ into $A$.
It should be remarked that sometimes it is possible to guarantee that we can continue even after choosing countably many points. In this case we have a linear order of an uncountable cofinality.
Finally, let me point out that everything above requires the axiom of choice. And it is consistent with the failure of the axiom of choice that there is an infinite linear ordering which has no countably infinite subset (so certainly not one which is isomorphic to either $\Bbb N$ or the negative integers). In particular we can arrange to have this linear order as a subset of the real numbers with their usual order.
Another easy proof, which can also be accounted for explaining this theorem, is that a linear order with no infinite decreasing chains is a well-order, and therefore has an initial segment isomorphic to $\Bbb N$.
So given a linear order, if there is a decreasing chain, it witnesses an embedding from the negative integers into the order; if there are no infinite decreasing chains, then this is a well-order and we're done.