Answer
Take, for example
$g(x)=x+1,\ \displaystyle \quad f(x)=\frac{1}{x-1}$.
This does not contradict Th.9.
Work Step by Step
Observe the hypotheses of theorem 9:
If $f$ is continuous at $c$ and $g$ is continuous at $f(c)$... then the composite function etc.
Select a continuous function $g(x)=x+1$
Note that g is continuous at x=0, $g(0)=1.$
Now, select f so that Theorem 9 does not apply.
We want f that is NOT defined at $x=g(0)=1$
Place $(x-1)$ in the denominator.
$f(x)=\displaystyle \frac{1}{x-1}$
Note that $f$ is continuous at x=0, $ f(0)=-1.$
The composition,
$f\displaystyle \circ g(x)=f[g(x)]=\frac{1}{g(x)-1}=\frac{1}{x+1-1}=\frac{1}{x}$,
is discontinuous at x=0.
No contradictions with Theorem 9, as it did not apply to this case.
