I've underlined the step in the proof that isn't clear to me.
I know what I said in the title is wrong; but I'm not really sure what motivates the step that I underlined.
This is taken from Royden, 4th edition, page 66.
Any thoughts would be greatly appreciated.
Thanks in advance.
Edit: To be more clear, I'm confused about how we get an open interval fully contained in $F_i$ which also contains $x$. Why must $F_i$ fully contain an open interval that also contains $x$?
To say that $g$ is continuous on $F$ means that for any $x \in F$ and for any $\epsilon>0$ there is some $\delta >0$ such that if $|y-x|<\delta$ and $y \in F $ then $|g(y)-g(x)| < \epsilon$.
In the above case, we have $x \in F_i$, and since the $F_k$ are disjoint and closed, the set $\cup_{k\ne i} F_k$ is closed and disjoint from $F_i$. Hence $x \notin \cup_{k\ne i} F_k$ and so there is some interval $(a,b)$ containing $x$ such that $(a,b) \cap (\cup_{k\ne i} F_k) = \emptyset$.
Hence if $y \in F \cap (a,b)$ we must have $y \in F_i$ and hence $g(y) = g(x)$.
So, given $x \in F$ and $\epsilon>0$, we can find some $i$ and an interval $(a,b)$ as above such that $x \in (a,b)$ and $(a,b) \cap F \subset F_i$ (in particular we have $g(x) = a_i$).
Now choose $\delta>0$ such that $B(x,\delta) \subset (a,b)$, then if $|y-x| < \delta$ and $y \in F$ we have $y \in (a,b) \cap F_i$ and so $g(y) = a_i$ (and so $|g(y)-g(x)| < \epsilon$, of course).
