Given a limit ordinal $\alpha$ that is not a cardinal, we can set $\kappa=|\alpha|$ and find a one-to-one sequence $\langle \alpha_\nu\mid\nu<\kappa\rangle$ with range $\alpha$. We want to rigorously construct an increasing subsequence with limit $\alpha$.
This question has been asked before, but I am having difficulties to show that the corresponding subsequence has limit $\alpha$.
I believe one should use the fact that $\alpha$ is a limit ordinal, but I don't see how to apply it.
Here is the argument in more detail.
Let $\eta_0=0$. Suppose that $\gamma<\kappa$, and that $\eta_\xi$ has been defined for $\xi<\gamma$ so that $\langle\alpha_{\eta_\xi}:\xi<\gamma\rangle$ is strictly increasing, and if
$$A_\xi=\{\nu<\kappa:\alpha_{\eta_\zeta}<\alpha_\nu\text{ for all }\zeta<\xi\}$$
for each $\xi<\gamma$, then
$$\eta_\xi=\min\{\nu\in A_\xi:\eta_\zeta<\nu\text{ for all }\zeta<\xi\}\,.\tag{1}$$
If $\sup_{\xi<\gamma}\alpha_{\eta_\xi}=\alpha$, we’re done. Otherwise, let
$$A_\gamma=\{\nu<\kappa:\alpha_{\eta_\xi}<\alpha_\nu\text{ for all }\xi<\gamma\}\ne\varnothing\,,$$
and let
$$A_\gamma'=\{\nu\in A_\gamma:\eta_\xi<\nu\text{ for all }\xi<\gamma\}\,.$$
If $A_\gamma'=\varnothing$, then for each $\nu\in A_\gamma$ there is a $\xi<\gamma$ such that $\nu<\eta_\xi$; this violates the minimality of $\eta_\xi$ in $(1)$, so $A_\gamma'\ne\varnothing$, and we set $\eta_\gamma=\min A_\gamma'$. Then $\langle\alpha_{\eta_\xi}:\xi\le\gamma\rangle$ is strictly increasing, and $(1)$ is satisfied for $\xi\le\gamma$, so the recursive construction goes through until at some stage $\gamma$ we have $\sup_{\xi<\gamma}\alpha_{\eta_\xi}=\alpha$, and at that point $\langle\alpha_{\eta_\xi}:\xi<\gamma\rangle$ is the desired subsequence.