Let $X$ and $M$ be a metric spaces.
Given $A, B$ compact subsets of $M$, the Hausdorff distance between them, wich we denote by $d_{H}(A,B) =$ max $\{ d_A(B), d_B(A) \}$, where
\begin{align} d_A(B) &= \sup_{b \in B} d(A,b) \\ d_B(A) &= \sup_{a \in A} d(B,a) \end{align}
Denote by $C(M)$ the space of compact subsets of $M$ endowed with the Hausdorff metric.
(Definition 1) We say that a map $\Gamma: X \to C(M)$ is lower semicontinuous in a point $x \in X$ if for every open set $U$ such $U \cap \Gamma(x) \neq \emptyset$ there exists a neighborhood $V$ of $x$ in $X$ such that if $y \in V$ then $U \cap \Gamma(y) \neq \emptyset$.
(Definition 2) Following this paper, they observe that a map is also lower semicontinuous if for any $\varepsilon >0$ there exists $\delta>0$ such that if $y \in B(x,\delta)$ then $d_{\Gamma(y)}(\Gamma(x)) < \varepsilon$.
I will prove that Definition 2 implies Definition 1.
Proof: Take $U$ an open set such that $U \cap \Gamma(x) \neq \emptyset$. Take $a \in U \cap \Gamma(x)$. There exists $\varepsilon > 0$ such that $B(a,\varepsilon) \subseteq U$. By hypothesis, there exists $\delta>0$ such that if $y \in B(x,\delta)$ then $d_{\Gamma(y)}(\Gamma(x)) < \frac{\varepsilon}{2}$.
We claim that $\Gamma(y) \cap B(a,\varepsilon) \neq \emptyset$.
Suppose, by contradiction, that for any $b \in \Gamma(y)$, one has $d(b,a) \geq \varepsilon$. So, $$\varepsilon \leq d(\Gamma(y),a) \leq \sup_{w \in \Gamma(x)} d(\Gamma(y),w) < \frac{\varepsilon}{2},$$
and we are done.
I have difficult to show that definition 1 proves definition 2.
Thanks in advance for your help.
By definition, $d_{\Gamma(y)}(\Gamma(x)) < \varepsilon$ if and only if $$ \sup_{a \in \Gamma(x)} \inf_{b \in \Gamma(y)} d(a,b) < \varepsilon, \tag{$\star$} $$ if and only if for every $a \in \Gamma(x)$ there exists $b \in \Gamma(y)$ such that $d(a,b) < \varepsilon$. Therefore, to show that Definition 1 implies Definition 2, we must show that for every $\varepsilon > 0$ there is $\delta > 0$ such that if $y \in B(x,\delta)$, then for each $a \in \Gamma(x)$ there is $b \in \Gamma(y)$ with $d(a, b) < \varepsilon$.
Now, for each $a \in \Gamma(x)$, it follows that $\Gamma(x) \cap B(a,\varepsilon) \ne \emptyset$, so by lower semicontinuity we can fix $\delta_a > 0$ such that $\Gamma(y) \cap B(a,\varepsilon) \ne \emptyset$ for all $y \in B(x,\delta_a)$. The sets $\{B(a,\varepsilon)\}_{a \in \Gamma(x)}$ form an open cover of the compact subset $\Gamma(x)$; hence, we may extract a finite sequence $a_1, \dots, a_n$ for which $\{B(a_j, \varepsilon)\}_{j =1}^{n}$ is a finite subcover. Let us take $\delta := \min \{\delta_{a_j} : 1 \leq j \leq n\} >0$. Then if $y \in B(x,\delta)$ and $a \in \Gamma(x)$ is arbitrary, we know that $a \in B(a_j, \varepsilon)$ for some $j=1, \dots, n$, and it follows that $\Gamma(y) \cap B(a_j, \varepsilon) \ne \emptyset$. Hence, there exists $b \in \Gamma(y)$ with $d(a,b) < \varepsilon$, so $d_{\Gamma(y)}(\Gamma(x)) < \varepsilon$, as desired.
By the way, looking at the inequality ($\star$) gives us a way to avoid a contradiction argument in your proof that Definition 2 implies Definition 1. Let's pick up from "We claim that $\Gamma(y) \cap B(a,\varepsilon) \ne \emptyset$." We actually don't need to use $\varepsilon/2$, as $\varepsilon$ is sufficient:
Since $d_{\Gamma(y)} (\Gamma(x)) < \varepsilon$, there exists $b \in \Gamma(y)$ such that $d(a,b) < \varepsilon$; whence $b \in \Gamma(y) \cap B(a,\varepsilon)$, as desired.
Finally, I just want to point out that Definition 2 is clearly stronger than Definition 1 if we relax the assumption that $\Gamma$ maps into compact sets.