Does continuity on $S$ implies continuity on $\partial S$?

Question

In the book of Analysis on Manifolds by Munkres, at page 110, it is given that

However, the function $f$ is continuous on $S$, and does not have be continuous on the boundary of $S$, so how can he argue as

"Continuity of $f_S$ at $x_0$ implies ..." ?

I mean we are already trying to show that $f_S$ is continuos at $x_0$, so I'm really confused how can he base his argument to the statement that he is trying to show.

Edit:

$f_X (x)= f(x)$ for $x \in X$, and $f_X(x) = 0$ for $x \not = X$.

Answer 1

No. Continuity on $S$ does not imply continuity on $\partial S$. Indeed, on $\mathbb{R}$, consider $f(x) = \frac{1}{x}$ on $B_{\frac{1}{2}}{(\frac{1}{2})}$ = (0, 1) := $S$. Then $f$ is continuous on $S$, but discontinuous on $\partial S$.

Edit: I didn’t notice that it stated $f$ is bounded. But as noted by @user571438, take $f(x)=\sin(1/x)$ on (0,1)