liminf and limsup properties First we introduce the following notation:
$$ 
\mathcal{N}_\infty:= \{N\subset \mathbb{N}| \mathbb{N} \text{\ }N \text{ is finite}\}
$$
and
$$ 
\mathcal{N}_\infty^\#:= \{N\subset \mathbb{N}| N \text{ is infinite}\}
$$
In most textbooks of real analysis, the limit inferior is defined in one of the following two ways:
$$
\liminf_n C_n = \bigcup_{n=1}^\infty \bigcap_{m=n}^\infty C_m
$$
or
$$
\liminf_n C_n = \left\{x \in \mathcal{X} | x\in C_k \text{ ultimately for all } k \right\}
$$
We need to show that:
$$
\liminf_n C_n = \bigcap_{N\in \mathcal{N}_\infty^\#} \overline{\bigcup_{n\in N}C_n}
$$
where the overline denotes the set closure in the respective topology. For the limit superior we need to show that:
$$
\limsup_n C_n = \bigcap_{N\in \mathcal{N}_\infty} \overline{\bigcup_{n\in N}C_n}
$$
These properties appear in [p.110, 1] as exercises.
[1] R.T. Rockafellar and R. J-B. Wets, "Variational Analysis", Grundlehren der mathematischen Wissenschaften, vol. 317, 1998.
 A: Thanks to Martin Sleziak (for pointing at the book of G. Beer [p.145, Prop. 5.2.2. in 1]) the proof is as follows:
Proposition 1. Let $(\mathcal{X},\mathcal{T})$ be a Hausdorff topological space. Then:
$$
\liminf_n C_n = \bigcup_{N\in\mathcal{N}_\infty^\#}\overline{\bigcap_{v\in N}C_v}
$$
Proof.
(1). Let $x\in\liminf_n C_n$ and let $\Sigma\in\mathcal{N}_\infty^\#$. Let $W$
be a neighborhood of $x$. There is a $N_0\in\mathbb{N}$ sucht that for all $n\geq N_0$
such that $n\in\Sigma$:
$$
W\cap C_n \neq \emptyset
$$
Thus,
$$
x\in\overline{\bigcup_{n\in\Sigma}C_n}
$$
(2). Assume that $x\notin \liminf_n C_n$. Then, there is an 
open neighborhood of $x$, let $W\ni x$, such that 
$\Sigma_0:=\{n\in\mathbb{N}| W\cap C_n = \emptyset\}$. Therefore,
$x\notin \overline{\bigcup_{n\in\Sigma_0}C_n}$. This completes the
proof. $\square$
Note 1: Characterization of the closure of a set: Let $C\subset \mathcal{X}$ and $\bar{C}$ denote its closure which is defined as:
$$
\bar{C}=\bigcap\{F\supset C| F^c\in \mathcal{T}\}
$$
Then:
$$
x\in\bar{C} \Leftrightarrow \forall V\in\mathcal{T},\ V\ni x:\ V\cap C \neq \emptyset
$$
Note 2: This result is stated in [1] for nets of sets in $\mathcal{X}$, $\{C_n\}_{n\in\Lambda}$ where $\Lambda$ is a partially ordered set. Then the class $\mathcal{N}_\infty^\#$ is replaced by the family of cofinal sets of $\Lambda$. Set set $\Sigma$ is called a cofinal subset of $\Lambda$ if for all $\lambda\in\Lambda,\ \exists\sigma\in\Sigma:\ \sigma\geq\lambda$.
Corollary 2. Let $\{C_n\}_{n\in\mathbb{N}}$ be a sequence of subsets of $\mathcal{X}$. Then:
$$
\bigcap_{n\in\mathbb{N}}C_n \subseteq \overline{\bigcap_{n\in\mathbb{N}}C_n} \subseteq \liminf_n C_n
$$
Proof. It follows from Proposition 1 taking $N=\mathbb{N}\in\mathcal{N}_\infty^\#$. $\square$
[1] G. Beer, "Topologies on Closed and Closed Convex Sets", Kluwer Academic Publishers, ISBN: 0-7923-2531-1.
A: I believe the overline you put shouldn't be there ; it has no reason to be. I'll prove what you want to prove without putting it there to prove my point.
I'll do the $\liminf$ case for you, the $\limsup$ case is symmetric. You want to show that
$$
\bigcup_{n=1}^{\infty} \left( \bigcap_{m =n}^{\infty} C_m \right) 
= 
\bigcap_{N \in \mathcal N_{\infty}^{\#}} \left( \bigcup_{n \in N} C_n \right).
$$
You were right wanting to say that you want to prove $(\subseteq)$ and $(\supseteq)$, but you can make your life more easier a little. To show that
$$
\bigcup_{n=1}^{\infty} \left( \bigcap_{m =n}^{\infty} C_m \right)
\subseteq
\bigcap_{N \in \mathcal N_{\infty}^{\#}} \left( \bigcup_{n \in N} C_n \right),
$$
you only need to show that 
$$
\forall N \in \mathcal N_{\infty}^{\#}, \qquad 
\bigcup_{n=1}^{\infty} \left( \bigcap_{m =n}^{\infty} C_m \right)
\subseteq 
\bigcup_{n \in N} C_n  
$$
since being in the intersection means being in all of the things you intersect over. Now suppose
$$
x \in \bigcup_{n=1}^{\infty} \left( \bigcap_{m =n}^{\infty} C_m \right) \qquad \Longrightarrow \qquad \exists n \, \text{ s.t. } x \in \bigcap_{m=n}^{\infty} C_m \qquad \Longrightarrow \qquad \forall m \ge n, \quad x \in C_m.
$$
Since $N$ is an infinite subset of $\mathbb N$, there exists $m \ge n$ with $m \in N$. Therefore there exists $m \in N$ such that 
$$
x \in C_m \subseteq \bigcup_{n \in N} C_n,
$$
hence we are done with this part.
It's just manipulations of large symbols, don't be astonished by the length of the proof. To show that 
$$
\bigcup_{n=1}^{\infty} \left( \bigcap_{m =n}^{\infty} C_m \right)
\supseteq
\bigcap_{N \in \mathcal N_{\infty}^{\#}} \left( \bigcup_{n \in N} C_n \right),
$$
You can restrict yourself to show that
$$
x \in \bigcap_{N \in \mathcal N_{\infty}^{\#}} \left(  \bigcup_{n \in N} C_n \right) \quad \Longrightarrow \quad \exists n \, \text{ s.t. } \,  x \in \bigcap_{m=n}^{\infty} C_m.
$$
Consider the set $N_X = \{ m \in \mathbb N \, | \, x \notin C_m \}$. If $N_X \in \mathcal N_{\infty}^{\#}$, we have
$$
x \in \bigcap_{N \in \mathcal N_{\infty}^{\#}} \left( \bigcup_{m \in N} C_m \right) \qquad \Longrightarrow \qquad x \in \bigcup_{m \in N_X} C_m,
$$
but this is a contradiction because $x$ is (by definition of $N_X$) not in any of those $C_m$'s. Thus $N_X \in N_{\infty}$, and therefore there exists $n \in \mathbb N$ such that for all $m \ge n$, $x \in C_m$. This completes the argument.
Hope that helps,
