Solution:

Note that, for every $x$, we can choose a rational number $h(x)$ such that $f(x)<h(x)<g(x)$. Proof: since $g(x)-f(x)>0$, we can choose an integer $n$ larger than $1/(g(x)-f(x))$. Then the open interval $(n f(x), n g(x))$ has width $n(g(x)-f(x))>1$, so it contains some integer $m$. So we have $f(x)<m/n<g(x)$, and we can take $h(x)=m/n$.

Now $h$ is a function mapping an uncountably infinite set, $\mathbf{R}$, to a countably infinite set, $\mathbf{Q}$ (the set of rational numbers). For any $z \in \mathbf{Q}$, consider $h^{-1}(z)$, the set of elements of $\mathbf{R}$ whose image under $h$ is $z$. Certainly $\mathbf{R}$ is the union of all such sets, since, for any $x \in \mathbf{R}$, $x$ lies in $h^{-1}(h(x))$. If the set $h^{-1}(z)$ is finite (or even countably infinite) for all $z \in \mathbf{Q}$, then $\mathbf{R}$ consists of a union of countably many countable sets, so it is countable. But this is false, so some $h^{-1}(z)$ must be uncountably infinite. Let $S=h^{-1}(z)$ accordingly. Then, for all $x, y \in S$, we have $f(x)<h(x)=z=h(y)<g(y)$, and $S$ is infinite, as needed.