Let's say I take a multifunction which goes like this:
$$f(x) = x^2 \chi_{x\le a} + (x^2 + 1) \chi_{x > a}$$
Let's say, I take $b > a$ and let's define $c := a^2$, $d_1 := (a^2 + 1)$ and $d_2 := (b^2 + 1)$
It's clear that
$$f((a,b)) = (d_1, d_2)$$
So this means, that
$$cl \ f((a,b)) = [ d_1, d_2 ]$$
It's also clear, that
$$f([a,b]) = \{ c \} \cup (d_1, d_2]$$
Function $f(x)$ is lower semi-continuous. So it should hold, that
$$f (cl \ (a,b)) \subset cl \ f ((a,b))$$
But it's not the case here!
$$\{c\} \cup (d_1, d_2] \not\subset [d_1, d_2]$$
So what's wrong here?
Yes, $f$ is lower semi-continuous. That means that, if $\tau$ is the usual topology on $\Bbb R$ and $\lambda$ is the topology on $\Bbb R$ generated by the intervals of the form $(a,\infty)$ ($a\in\Bbb R$), then $f$ is continuous as a function from $(\Bbb R,\tau)$ into $(\Bbb R,\lambda)$. And, in $(\Bbb R,\lambda)$,$$\overline{f\bigl((a,b)\bigr)}=\overline{(d_1,d_2)}=(-\infty,d_2].$$Since $c\in(-\infty,d_2]$, there is no contradiction here.