One can quite easily prove that if $(A,<)$ is linearly ordered and for all $a\in A$, $$|\{b\in A\mid b\leq a\}|<\aleph_\alpha,\quad (*)$$ then $|A|\leq \aleph_\alpha$.
The strict inequality in $(*)$ is essential for the proof. I'm wondering if there is an easy counter-example to this statement when $<$ has been replaced by $\leq$ in $(*)$?
If $A=\mathbb R, \mathbb Q,\mathbb N$, a counter-example cannot be found.
Let $A=\omega_1$ - the first uncountable ordinal. Then, for every $a\in A$, $a$ is countable, so the sets $\{b\in A\mid b\le a\}=\{a\}\cup a$ are countable as well, i.e. $|\{b\in A\mid b\le a\}|=\aleph_0$, however $|A|=\aleph_1$.