Glued along the rationals and the irrationals

($\Rightarrow$) Suppose $h$ is continuous at $x_0$. Take rationals $r_n\to x_0$ and irrationals $s_n\to x_0$ (possible by density). Then $h(r_n)=f(r_n)\to f(x_0)$ and $h(s_n)=g(s_n)\to g(x_0)$. By continuity of $h$ at $x_0$, both sequences $h(r_n)$ and $h(s_n)$ tend to $h(x_0)$. So $f(x_0)=h(x_0)=g(x_0)$.

($\Leftarrow$) Suppose $f(x_0)=g(x_0)=:\ell$. Then $h(x_0)=\ell$ whichever kind of number $x_0$ is. Let $\varepsilon>0$. By continuity of $f$ and $g$ at $x_0$ there exist $\delta_1,\delta_2>0$ with $|f(x)-\ell|<\varepsilon$ when $|x-x_0|<\delta_1$, and $|g(x)-\ell|<\varepsilon$ when $|x-x_0|<\delta_2$. With $\delta=\min(\delta_1,\delta_2)$, for $|x-x_0|<\delta$ the value $h(x)$ is either $f(x)$ or $g(x)$, so $|h(x)-\ell|<\varepsilon$.

Example. If $f(x)=x$ and $g(x)=x^2$, then $h$ is continuous exactly at $0$ and $1$, where $x=x^2$.