First, you might also want to take a look at this answer to a similar question.
Okay: the first description assumes that there is some sort of notion of "accumulation point" at work in the set $X$, as you surmise; this may be derived from a topology.
The second description talks about limit points, but you can apply it to any set by endowing the set with the discrete topology (every subset is open, every subset is closed). If you do that, then the definition is the usual definition of limit superior of a sequence of sets: it is the collection of all points that are in infinitely many of the terms of the sequence, while the limit inferior is the collection of all points that are in all sufficiently large terms of the sequence.
The "second way" of defining it is in terms of unions and intersection. If $\{X_n\}_{n\in\mathbb{N}}$ is a family of sets, then
\begin{align*}
\limsup_{n\in\mathbb{N}} X_n &= \bigcap_{n=1}^{\infty}\left(\bigcup_{j=n}^{\infty} X_j\right)\\\
\liminf_{n\in\mathbb{N}} X_n &= \bigcup_{n=1}^{\infty}\left(\bigcap_{j=n}^{\infty} X_j\right).
\end{align*}
This coincides with the notion of the limit superior being the set of all limit points of infinitely many terms in the sequence, under the discrete topology; and the limit inferior being the set of all limit points of all sufficiently large-indexed terms of the sequence (again, under the discrete topology).
The notion of "accumulation point" in the first description is more informal. If you are working with a topological space, then it is limit points as described above and by "accumulation set" you should read "set of all limit points".
For your third point, in order to be able to talk about joins and meets you need to have some sort of complete lattice order on your set, so that you can talk about those infinite meets and infinite joins; this is the case, for instance, in the real numbers; appropriately interpreted, you do get essentially the definition you propose, though you need to tweak it a bit in order to actually get what the actual definition is (see the other answer quoted above); you don't actually work with the points themselves, but with a slightly different set determined by the points.
I think that the previous answer linked to answers essentially your fourth point, of how to interpret limit superior and limit inferior of a sequence of points as a special case of limit superior and limit inferior of sets; but if this is not the case, point it out and I'll try to answer it de nuovo.
First let’s clear up a misconception: there is no last term in the expression for $B$. For any term $A_n \cap A_{n+1} \cap A_{n+2} \cap \dots$ there is a next term $A_{n+1} \cap A_{n+2} \cap A_{n+3} \cap\dots$ that is potentially a larger set.
Now to the problem itself: you must start with the definitions of $\limsup$ and $\liminf$ for sequences of functions. If $\langle f_n \rangle_n$ is a sequence of real-valued functions, $$\limsup_{n\to\infty} f_n(x) = \lim_{n\to\infty}\;\; \sup_{m\ge n}f_m(x),$$ and $$\liminf_{n\to\infty}f_n(x) = \lim_{n\to\infty}\;\;\inf_{m\ge n}f_m(x).$$ In your case $f_n = \chi_{A_n}$, so you’re trying to show that $$\lim_{n\to\infty}\;\;\inf_{m\ge n}\chi_{A_m}(x) = \chi_B(x)$$ and $$\lim_{n\to\infty}\;\; \sup_{m\ge n}\chi_{A_m}(x) = \chi_C(x)\;.$$
Since the only values of a characteristic function are $0$ and $1$, it should be clear that $\inf_{m\ge n}\chi_{A_m}(x)$ is always $0$ or $1$, and it’s not hard to see that it’s $1$ iff $\chi_{A_m}(x)=1$ for every $m\ge n$, i.e., iff $x\in A_m$ for every $m\ge n$. Similarly, $\sup_{m\ge n}\chi_{A_m}(x) = 0$ iff $\chi_{A_m}(x)=0$ for every $m\ge n$, which is the case iff $x \notin A_m$ for every $m\ge n$. From here it’s fairly straightforward to get the desired results.
Best Answer
I must admit that I did not know these definitions, either.
Yes, because if $x \in \liminf A_n$ you have a sequence $\{x_k\}$ with $x_k \in A_k$ and $x_k \rightarrow x$ and you can choose your subsequence $\{A_{n_k}\}$ to be your whole sequence $\{A_n\}$.
For $X = \mathbb{R}$ take $A_n = \{0\}$ for all $n$. Then you have $(\liminf A_n)^c = \{0\}^c = \mathbb{R}\setminus{\{0\}}$, but $\limsup A_n^c = \mathbb{R}$. So in general you do not have equality.
Of course you have "$\supseteq$". But if you take $A_n := [-1,1- \frac{1}{n}) \subseteq \mathbb{R}$ you have the sequence $\{x_k\}$ defined by $x_k := 1 - \frac{2}{k}$ which converges to $1$ which is not in $\overline{\bigcap_{n = m}^{\infty} A_n} = [-1,1-1/m]$ for any $m \in \mathbb{N}$. But it works of course, if $\{A_k\}$ is decreasing.
In your first assumption you have "$\subseteq$". A sequence $\{x_k\}$ with $x_k \in A_{n_k}$ with $\{A_{n_k}\}$ some subsequence of $\{A_n\}$ lies eventually in $\bigcup_{n = m}^\infty A_n$ for all $m$ by the definition of subsequence. Now for metric spaces you have "$\supset$", too, as you can start by choosing $x_1 \in A_{n_1}$ with $d(x_1,x) < 1$ and proceed by induction choosing $x_{n_k} \in A_{n_k}$ with $n_k > n_{k-1}$ and $d(x_{n_k},x) < \frac{1}{k}$. I doubt that this holds in general (at least this way of proving it does not).