Note. This question concerns steps in a specific proof of the statement from the title, a link to which is (was) provided in the question. That link is now dead, and since the question does not contain all of the information required to accurately answer the question, it will be closed.
I'm reading some notes with a proof that a first countable, countably compact space is sequentially compact.
On page 118 of these notes (third page of the pdf), they are constructing a convergent subsequence. They state that $B_{n(k)+1}\cap T_{n(k)+1}\neq\emptyset$, so you can take an $x_{n(k+1)}\in B_{n(k)+1}$, showing there exists an $x_{n(i)}\in B_{n(i)}$ for all $i\in\mathbb{N}$.
How do they get that? Don't you only have that $x_{n(k+1)}\in B_{n(k)+1}$, not $B_{n(k+1)}$?