Upper and lower hemicontinuity of $C$ implies that $C$ is continuous in the Hausdorff metric, $d_H$, considering $C: X \to P(X)$, where $P(X)$ denotes the set of all closed subsets of $X$.
As such, $C(x_n) \to C(x)$ where convergence is wrt to $d_H$.
This answer provides a partial answer: $\lambda$ is upper semicontinuous wrt $d_H$, that is $\lim_{x_n \to x}\lambda(C(x_n))\leq \lambda(C(x))$.
If $C$ is convex-valued we have that the inequality holds with equality.
A sketch of the proof is as follows:
If $C(x)$ is a singleton, we are done, as $0=\lambda(C(x))$ and by upper semicontinuity, $\lambda(C(x))=0\geq \lim_{x_n \to x}\lambda(C(x_n))\geq 0$.
If $C(x)$ has a non-empty interior, then $\lambda(C(x))>0$.
Thus, recalling that $C$ is convex-valued, this implies that $\exists N$ such that $\forall n\geq N$, $C(x_n)$ also has a nonempty interior.
Let $f_n(y):=1_{y \in \text{int}(C(x_n))}$.
Note that $f_n$ is measurable with respect to the Lebesgue measure on $(X,\mathcal B)$ and dominated by 1 for all $n$.
To see that $f_n$ converges pointwise to $f$, where $f(y)=1_{y \in \text{int}(C(x))}$, take any $y \in \text{int}(C(x))$. Let $\epsilon:=\min_{z \in \partial C(x)}\lVert z-y\rVert_\infty>0$.
Let $B_{\delta}(z)$ be the $\delta$ open-ball around $z\in X$ wrt the sup-norm.
Then, $B_{\epsilon/2}(y)\subset \text{int}(C(x))$ and as $d_H(C(x_n),C(x))\to 0 \implies d_H(\partial C(x_n),\partial C(x))\to 0$, we have that $\exists N'\geq N$ such that $\forall n \geq N'$ where $B_{\epsilon/2}(y) \subset co(\partial C(x_n))=C(x_n)$ (as $C$ is convex-valued), where $\partial A$ denotes the boundary of set $A$ and $co(A)$ its convex hull.
Then, by Lebesgue's dominated convergence theorem we have that
$\lambda(C(x_n))=\lambda(\text{int}(C(x_n)))=\lim_{n\to \infty}\int_X f_n d\lambda = \int_X \lim_{n\to \infty} f_n d\lambda = \int_X f d\lambda = \lambda(\text{int}(C(x))) = \lambda(C(x)).$
$\newcommand{\gr}{\operatorname{Gr}}$ It is not true in general that $\gr(\Gamma)$ is path-connected. Consider: $$\Gamma:[0,1]\ni x\mapsto\begin{cases}[-1,1]&x=0\\\{\sin x^{-1}\}&0<x<1\end{cases}\subseteq\Bbb R$$Then $\gr(\Gamma)$ is the closed topologist's sine curve, which is well known to be connected but not path connected (nor locally connected nor locally path connected). We should also check $\Gamma$ is complex and convex valued as well as upper hemicontinuous, but this is nearly obvious.
So, why should $\gr(\Gamma)$ be connected? As commented, I don't like your proof since the convergent sequence $(x_{n_k})_{k\in\Bbb N}$ comes from nowhere. Firstly let's note that:
$\Gamma:[0,1]\to\Bbb R$ is compact and convex valued
Is a roundabout way of saying: $\Gamma(x)$ is a closed and bounded interval for every $x$. By a simple compactness argument (utilising upper hemicontinuity) we can conclude $\gr(\Gamma)$ is bounded. Using this closed graph theorem, $\gr(\Gamma)$ is then compact.
Suppose there exists a disconnection of $\gr(\Gamma)$ into $A$ and $B$. It follows that $A,B$ are both compact. Let $\pi:[0,1]\times\Bbb R\to[0,1]$ be the projection: then, $\pi A$ and $\pi B$ are both compact hence closed. Evidently $\pi A\cup\pi B=[0,1]$ with neither set being empty. Note that $\Gamma(x)$ is connected for all $x$, so $\{x\}\times\Gamma(x)$ must be fully contained in either $A$ or $B$ for any $x$. It follows that $\pi A\cap\pi B=\emptyset$. We conclude that $\pi A$ and $\pi B$ form a disconnection of $[0,1]$, contradicting the fact that $[0,1]$ is connected.
Therefore no $A,B$ can exist; therefore $\gr(\Gamma)$ is connected.
Generalising, if $\Gamma:X\to Y$ is connected-valued and $X$ is a connected Hausdorff space, then if $\gr(\Gamma)$ is compact we get to conclude $\gr(\Gamma)$ is connected.
Best Answer
I beleive I got it. The mapping $f \colon \mathbb{R} \to \mathcal{P} (\mathbb{R})$ given by $f(x)=[-1,1]$ for $x \leq 0$ and by $f(x)= [-1,1] \setminus \Big\lbrace t \in \mathbb{R} \, ; \ \big|t-\sin \frac{1}{x} \big|< \frac{x}{x+1} \Big\rbrace$ for $x > 0$ seems to work fine.