In general there is no continuous function between u.s.c. $f$ and l.s.c. $g\leq f$. For example, take $f(x)=1, 0\leq x\leq 1;\; f(x)=0, 1<x\leq 2$, this is u.s.c. Now $g(x)=1, 0\leq x<1;\; g(x)=0, 1\leq x\leq 2$, then $g(x)\leq f(x)$ and evidently there is no continuous function in between.

In general there is no continuous function between u.s.c. $f$ and l.s.c. $g\leq f$. For example, take $f(x)=1, 0\leq x\leq 1;\; f(x)=0, 1<x\leq 2$, this is u.s.c. Now $g(x)=1, 0\leq x<1;\; g(x)=0, 1\leq x\leq 2$, this is l.c.s, and $g(x)\leq f(x)$ and evidently there is no continuous function in between.

Moreover, it is easy to arrange a decreasing sequence of continuous function tending to $f$ from above, and increasing sequence of continuous functions tending to $g$ from below. This the answer to your question is no, even if your functions $f_n$, $g_n$ are continuous.