How to show that a set-valued function is lower semicontinuous? Let me recall the definitions of lower and upper semicontinuity of real-valued functions.
Definition 1.
Let $X$ be a topological space and $f:X→\mathbb R$. We say that $f$ is lower semicontinuous (resp.: upper semicontinuous), if the set $\{x\in X: f(x)>a\}$ $\left(\text{resp.} :\{x\in X: f(x)<a\}\right)$ is open for every $a\in\mathbb R$. 
For set-valued maps, the definition of lower semicontinuity reads as follows.
Definition 2. 
Let $X,Y$ be topological spaces and $\Phi:X\to \mathcal P(Y)$ be a set-valued mapping. We say that $\Phi$ is lower semicontinuous, if for every open set $G\subset Y$, the set $\{x\in X:\Phi(x)\cap G\neq\emptyset\}$ is open in $X$.
Question
Now let $f_1,f_2:X\to\mathbb R$, $f_1\leq f_2$, where $f_1$ is upper semicontinuous and $f_2$ is lower semicontinuous. How to show that a set-valued map $x\mapsto [f_1(x),f_2(x)]$ is lower semicontinuous?
 A: The key observation is this, defining $F(x)=[f_(x),f_2(x)]$ and considering any open interval $(a,b) \subseteq \Bbb R$
$$F(x) \cap (a,b) \neq \emptyset \iff (f_1(x) < b) \land (f_2(x) > a) \tag{1}$$

($\Rightarrow$) suppose $y \in [f_1(x), f_2(x)] \cap (a,b)$. Then $f_2(x) \ge y >  b$ and $f_1(x) \le y < b$ so the right hand side holds. 
  ($\Leftarrow$) Suppose that $F(x) \cap (a,b)=\emptyset$, As both sets are intervals, this means that one of them lies completely left of the other one, so $[f_1(x),f_2(x)]$ lies to the left of $(a,b)$, in which case $f_2(x) \le a$ contradicting the assumption on the right, or the other way round, but then $f_1(x) \ge b$, contradiction again. 

This means that the set $$F^{\Leftarrow}[(a,b)]:=\{x: F(x) \cap (a,b)\neq \emptyset \} = f_1^{-1}[(\leftarrow,b)] \cap f_2^{-1}[(a,\rightarrow)]$$
is open as a finite intersection of open sets (as $f_2$ is lsc and $f_1$ is usc). And if $O$ is any open subset of $\Bbb R$, it is a union of open intervals $(a_i, b_i)$ and 
$$F^{\Leftarrow}[O] = F^{\Leftarrow}[\bigcup_i (a_i, b_i)]= \bigcup_i F^{\Leftarrow}[(a_i,b_i)]$$
(the last identity is easily checkable from the definitions) and so is open as a union of open sets. This shows $F$ is lsc as a set map. 
