If every initial segment of the order has only countably many predecessors, then every countable subset of the order would be bounded (for otherwise the whole order would be a countable union of countable sets). In this case, we may therefore construct an increasing $\omega_1$ sequence in the order by recursion: start by choosing any point. After any countable number of stages, you've picked at most countably many points, which are bounded, and so there remains points above for you to pick. Thus, by recursion, you find an increasing $\omega_1$ sequence from the order.

That

In case anyone objects to the fact that the previous argument used uses the axim axiom of choice, and this let me point out that one cannot prove the result without any use is necessary, in view of the followingaxiom of choice. It To see this, recall that it is known to be relatively consistent with ZF that $\omega_1$ is singular, the limit of countably many countable ordinals. Suppose that $\omega_1=\sup_n\alpha_n$, where $\alpha_0\lt\alpha_1\lt\cdots$ and so on. Let $L$ be the linear order obtained by chopping $\omega_1$ into the $\omega$ many blocks corresponding intervals $[\alpha_n,\alpha_{n+1})$, but swapping turning each of these blocks individually intervals upside down, to have the reverse order. So we have an $\omega$ increasing sequence of blocks, each is anti-well-ordered with countable order type. This is an uncountable order, and every initial segment is countable, because it is a finite union of countable sets. But there is no increasing $\omega_1$ sequence , in $L$, because any well-ordered sequence can have at most finitely many members of each (anti-well-ordered block), and there are $\omega$ many blocks in order. So every increasing suborder has finite order type or order type $\omega$ at most. Thus, AC it is required for consistent with ZF that the desired result about orders is false.

2 AC required

If every initial segment of the order has only countably many predecessors, then every countable subset of the order would be bounded (for otherwise the whole order would be a countable union of countable sets). In this case, we may therefore construct an increasing $\omega_1$ sequence in the order by recursion: start by choosing any point. After any countable number of stages, you've picked at most countably many points, which are bounded, and so there remains points above for you to pick. Thus, by recursion, you find an increasing $\omega_1$ sequence from the order.

That argument used the axim of choice, and this use is necessary, in view of the following. It is consistent with ZF that $\omega_1$ is singular, the limit of countably many countable ordinals. Suppose that $\omega_1=\sup_n\alpha_n$, where $\alpha_0\lt\alpha_1\lt\cdots$ and so on. Let $L$ be the linear order obtained by chopping $\omega_1$ into $\omega$ many blocks $[\alpha_n,\alpha_{n+1})$, but swapping these blocks individually to have the reverse order. So we have an $\omega$ increasing sequence of blocks, each is anti-well-ordered with countable order type. This is an uncountable order, and every initial segment is countable. But there is no increasing $\omega_1$ sequence, because any well-ordered sequence can have at most finitely many members of each (anti-well-ordered block), and there are $\omega$ many blocks in order. So every increasing suborder has finite order type or order type $\omega$ at most. Thus, AC is required for the result.

1

If every initial segment of the order has only countably many predecessors, then every countable subset of the order would be bounded (for otherwise the whole order would be a countable union of countable sets). In this case, we may therefore construct an increasing $\omega_1$ sequence in the order by recursion: start by choosing any point. After any countable number of stages, you've picked at most countably many points, which are bounded, and so there remains points above for you to pick. Thus, by recursion, you find an increasing $\omega_1$ sequence from the order.