Solutions

a) Suppose that $g$ and $h$ are continuous at $a$, and that $g(a)=h(a)=l$. Pick any $\varepsilon>0$. By definition, there exist $\delta_1$ and $\delta_2$ such that

$$|x-a|<\delta_1 \implies |g(x)-l|<\varepsilon$$

and

$$|x-a|<\delta_2 \implies |h(x)-l|<\varepsilon.$$

Hence, with $\delta=\min(\delta_1,\delta_2)$, we have

$$0\leq x-a<\delta \implies |g(x)-l|<\varepsilon$$

and

$$0<a-x<\delta \implies |h(x)-l|<\varepsilon.$$

Since $f(a)=g(a)$, $f$ is continuous at $a$.

b) For $x\in [a,b)$, $f$ is continuous because $f(x)=g(x)$ and $g$ is continuous over that interval. Likewise, for $x\in(b,c]$, $f$ is continuous over that interval because $f(x)=h(x)$ and $h$ is continuous over it. For $x=b$, $f$ is continuous because of the result proved in a).