Given a well order $(W,\le)$, where W is uncountable, and $\omega_1:= \{x\in W:$ only countably many $y \in X$ s.t. $y \le x\}$, prove $\omega_1$ is first countable.
I saw a proof saying that $\{(a,x]: a < x\}$ is a local basis at $x$, if $x$ is not the least element of $\omega_1$, but I haven't figured out why each element of the local basis is open.
Hint: Given any $y\in\omega_1,$ let $S(y)$ be the least element of $\omega_1$ greater than $y.$ (You may need to prove that such an element exists.) Now, consider the set of intervals $\bigl(x,S(y)\bigr)$ with $x<y$ if $y$ is a non-least element of $\omega_1.$ On the other hand, if $y$ is the least element of $\omega_1,$ then $\{y\}$ is open (why?), and so we can conclude that...?