$\limsup $ and $\liminf$ of a sequence of subsets relative to a topology
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).
First of all, I'd like to notice that the closure of a set $A \subset X$ is NOT, in general, the same as the set of all limit points (for sequences) of $A$. Nevertheless, given the topology $\tau$ on $X$, you can always define another topology where the closed sets are those that contains all its limit points (for sequences). This new topology is stronger (or equal) the original one, since the closed sets contains all its limit points. Let's assume our topology is of this type!
Regarding the notions for $\limsup$ and $\liminf$, notice that whenever you have a countably complete lattice you have the $\limsup$ and $\liminf$ for sequences. That is, if you have an order where every "growing" sequence of elements $x_n$ have a limit: $\sup x_n$. Similarly, every "decreasing" sequence of elements $x_n$ have a limit: $\inf x_n$. If you notice that $y_n = \inf_{j \geq n} x_n$ is a "growing" sequence, then, you will realise that it does have a limit: $\lim y_n = \sup_n \inf_{j \geq n} x_n$ This is the $\liminf x_n$. That is, $$ \liminf x_n = \sup_{n}\, \inf_{j \geq n} x_j. $$ Analogously, $$ \limsup x_n = \inf_{n}\, \sup_{j \geq n} x_j. $$ You say that $x_n$ has a limit when $\limsup x_n = \liminf x_n$.
One of the lattices one might be interested is the power set of $X$: $\mathcal{P} = \mathcal{P}(X)$. That is, the family of all subsets of $X$. This is a lattice where $\inf A_\lambda = \bigcap A_\lambda$ and $\sup A_\lambda = \bigcup A_\lambda$. Here, I use $\lambda$ to emphasise that the family might not be countable.
If you have a $\mathcal{L} \subset \mathcal{P}$, it might happen that for $A_n \in \mathcal{L}$, $\sup A_n \not \in \mathcal{L}$, or $\inf A_n \not \in \mathcal{L}$. However, if $\inf A_\lambda \in \mathcal{L}$ for every family $A_\lambda$, then we still have a supremum in $\mathcal{L}$ for $A_n$. To avoid confusion, let's denote the supremum and infimum in $\mathcal{L}$ by $\bigvee$ and $\bigwedge$. That is, $$ \bigwedge A_n = \inf A_n, $$ and $$ \bigvee A_n = \inf_{\substack{A \in \mathcal{L}\\A \supset \sup A_n}} A_n. $$ This is what we do if we have a family of groups, or a family of topologies, for example. If $X = G$ is a group, then the intersection of any family of subgroups is a group. But the union is not. The group generated by $H_\lambda \subset G$ is not $\sup H_\lambda$, but rather, $\bigvee H_\lambda$; that is, the smallest subgroup that contains all $H_\lambda$. Notice that the "generated group" could be determined for an arbitrary family of sets $S_\lambda$ instead of $H_\lambda$. But notice that $\bigvee S_\lambda$ would be just the same as $\bigvee \langle S_\lambda \rangle$, where $\langle S_\lambda \rangle = \bigwedge_{H \supset S_\lambda} H$. Nevertheless, there is a problem if we want to define $\bigwedge S_\lambda$. We could say that it is the smallest group that contains $\bigcap S_\lambda$, that is, $$ \bigwedge S_\lambda = \bigwedge_{H \supset \bigcap S_\lambda} H; $$ or, we could say that it is $$ \bigwedge S_\lambda = \bigwedge \langle S_\lambda \rangle. $$ Those sets might not be the same! (why?) But they will be the same if $S_\lambda$ is a subgroup. That is, if $S_\lambda = \langle S_\lambda \rangle$. This ambiguity also makes it ambiguous to talk about the $\liminf$, but not about $\limsup$ for $S_\lambda$ in the lattice of subgroups of $G$. (why?)
Now, instead of a group, let's consider the family given by the closed sets mentioned at the beginning of this answer. Given $A \subset X$, consider $\langle A \rangle = \overline{A}$. That is, $\langle \cdot \rangle$ is simply the closure operation, or, the set of all limit points of sequences in $A$. Then, for this lattice of closed sets, we can define $$ \begin{align*} \liminf(1) A_n &= \liminf \overline{A}_n \\&= \bigvee_{n} \bigwedge_{j \geq n} \overline{A}_j \\&= \bigvee_{n} \bigcap_{j \geq n} \overline{A}_j \\&= \overline{\bigcup_{n} \bigcap_{j \geq n} \overline{A}_j}. \end{align*} $$ But we could also define $$ \begin{align*} \liminf(2) A_n &= \bigvee_{n} \bigwedge_{F \supset \cap_{j \geq n} A_j} F \\&= \bigvee_{n} \overline{\bigcap_{j \geq n} A_j} \\&= \overline{\bigcup_{n} \overline{\bigcap_{j \geq n} A_j}} \\&= \overline{\bigcup_{n} \bigcap_{j \geq n} A_j}. \end{align*} $$ These two are not the same! But it is clear that $\liminf(2) A_n \subset \liminf(1) A_n$.
Now, notice that in the definition of $\liminf$ in your question, $x \in \liminf A_n$ iff for any neighbourhood $V$ of $x$, you have that $V \cap A_j$ for all $j \geq n$ for a certain $n$. That happens iff for some $n$, $V \cap \bigcap_{j \geq n} \overline{A}_j \neq \emptyset$. And this just means that $$ x \in \overline{\bigcup_{n} \bigcap_{j \geq n} \overline{A}_j}. $$ That is, the $\liminf$ in the question is exactly $\liminf(1)$.
Using the same argument, one can prove that the $\limsup$ in your question is the same as the $\limsup$ in this answer.
Notice that your statement "2" is not correct because while $\limsup$ is a closed set, the complement of $\liminf$ is an open set! Also, you could do the same construction with the "open sets". This would yield a not ambiguous $\liminf$ and an ambiguous $\limsup$. I guess that taking $\limsup$ for the closed sets and $\liminf$ for the open sets would yield the validity of statement "2". I think this is what you want in statement "4", when you use the "interior" of a set.
Edit: Fixed definition of $\langle \cdot \rangle$.