Important note: According to @Did 's comments, I need to clarify that in my original question I confused two different definitions of limit for sequences of sets. The limit defined as $\liminf_n C_n = \left\{x \in \mathcal{X} | x\in C_k \text{ ultimately for all } k \right\}$ is different from the inner limit of Rockafellar and Wets. In what follows I use the definition given by Rockafellar and Wets.
Notation: Let $\left( \mathcal{X},\mathcal{T}\ \right)$ be a topological space. We denote the family of open neighbourhoods of $x\in\mathcal{X}$ by $\mho(x):=\{V\in\mathcal{T}:\ x\in V\}$.
Proposition 1: Let $\{C_n\}_{n\in\mathbb{N}}$ be a sequence of sets in a Hausdorff topological space $\left( \mathcal{X},\mathcal{T}\ \right)$. Then, $ \liminf_n C_n = \{x|\forall V\in\mho(x),\ \exists N\in \mathcal{N}_\infty, \forall n\in N: C_n\cap V\neq \emptyset\} $ or equivalenty: $ \liminf_n C_n = \{ x|\forall V\in\mho(x),\ \exists N_0\in \mathbb{N}, \forall n\geq N_0: C_n\cap V\neq \emptyset \} $
Proof.
(1). If $x\in\liminf_n C_n$ then we can find a sequence $\{x_k\}_{k\in\mathbb{N}}$ such that $x_k\to x$ while $x_k\in C_{n_k}$ and $\{n_k\}_{k\in\mathbb{N}}\subseteq \mathbb{N}$ is a strictly increasing sequence of indices. For any $V\in\mho(x)$ there is a $N_0\in\mathbb{N}$ such that for all $i\geq N_0$ it is: $x_i\in V$; but also $x_i\in C_{n_i}$. Thus $C_{n_i}\cap V\neq \emptyset$. Therefore $x$ is in the right-hand side set of the equation.
(2). For the reverse direction assume that $x$ belongs to the right-hand side set of given equation. Then, there is a strictly increasing sequence $\{n_k\}_{k\in\mathbb{N}}$. Then, for every $V\in\mho(x)$ we can find a $x_k\in C_{n_k}\cap V$. Hence, $x_k\to x$ ( in the topology $\mathcal{T}$ ).
$\square$
Proposition 2: Let $(\mathcal{X},\|\cdot\|)$ be a normed space and $\{C_n\}_{n\in\mathbb{N}}$ be a sequence of sets in $\mathcal{X}$. The inner limit of a sequence of sets is: $ \liminf_n C_n = \{ x\in X | \lim_n d(x,C_n)=0 \} $
Proof.
(1). We now need to show that $\limsup_n d(x,C_n)=0$. Let us assume that $\limsup_n d(x,C_n)>0$, i.e. there exists an increasing sequence of indices $\{n_k\}_{k\in\mathbb{N}}$ so that $d(x,C_{n_k})\to_k a > 0$. This suggests that there is a $\varepsilon_0>0$ such that for all $k\in\mathbb{N}$ one has that $d(x,C_{n_k})>\varepsilon_0$. However, according to proposition \ref{propo:un_int}, $x\in\text{cl}\bigcup_{k\in\mathbb{N}}C_{n_k}$ while $d(x,\text{cl}\bigcup_{k\in\mathbb{N}}C_{n_k})\geq\varepsilon_0$ which is a contradiction. Hence, $\limsup_n d(x,C_n)=0$, i.e. $\lim_n d(x,C_n)=0$ and this way we have proven that $x$ is in the right-hand side set.
(2). Assume that $x$ in the right-hand side of the given equation. This is $\lim_n d(x,C_n)=0$. For any $\varepsilon>0$, we can find $n_0\in\mathbb{N}$ such that $d(x,C_{n})\leq \frac{\varepsilon}{2}$ for all $n\geq n_0$. By definition, we have that $d(x,C_{n})=\inf\{\|x-y\|,\ y\in C_{n}\}$, thus we can find a $y_n\in C_{n}$ such that
$\|y_n-x\|
we can do that following the steps pointed out by Davide Giraudo in his answer to this question. That is:
$ \exists\ y_n\in C_{n}:\ \|y_n-x\|<\varepsilon $ Therefore, $x\in C_{n} + \varepsilon \mathcal{B}$ from which it follows that $x\in\liminf_n C_n$ (According to proposition 1).
$\square$
Proposition 3: Let $(\mathcal{X},\|\cdot\|)$ be a normed space and $\{C_n\}_{n\in\mathbb{N}}$ be a sequence of sets in $\mathcal{X}$. The inner limit of a sequence of sets is: $ \limsup_n C_n = \{ x\in X | \liminf_n d(x,C_n)=0 \} $
Note: We know that $\liminf_n C_n \subseteq \limsup_n C_n$. We may therefore prove that :
$ \liminf_n C_n \subseteq \{ x\in X | \liminf_n d(x,C_n)=0 \} $