Attempting to find an unified treatment of the separation axioms Disclaimer
This question has been once edited to incorporate the suggestion of the answers given below but that post too contained gaps and so I have decided to rollback the post to its second version.
Background
Motivated by this answer, I have been trying to find an answer to the question that I have asked here. This post is an attempt to answer the question. For the definition of the spaces with which I have dealt see this page. 
Note that even though I write the conditions of being a $T_i$-space symbollicaly, don't take the expressions to be a formula of some formal language. Just read them as "convenient abbreviation" of the intended English language sentence.

Definition. Let $(X,\tau)$ be a topological space. Then for each $A\subseteq X$ we define, $$\mathcal{N}_A:=\{U:A\subseteq U\land U\in \tau\}$$and $$\mathcal{C}_A:=\{V:A\subseteq V\land X\setminus V\in\tau\land (\exists U\in\mathcal{N}_A)(A\subseteq U\subseteq V)\}$$If $A=\{x\}$ then instead of $\mathcal{N}_{\{x\}}$ and $\mathcal{C}_{\{x\}}$ we will simply write $\mathcal{N}_{x}$ and $\mathcal{C}_{x}$.
$T_0$-Space

Theorem 1. A topological space $(X,\tau)$ is $T_0$ iff $$\forall x\forall y((\{x\}\not\subseteq \{y\}\land \{y\}\not\subseteq \{x\})\iff(\mathcal{N}_x\not\subseteq\mathcal{N}_y\lor \mathcal{N}_y\not\subseteq\mathcal{N}_x))$$
Proof. Let $x\ne y$ and let $X$ be $T_0$. Then since $X$ is $T_0$, by the definition of a $T_0$-space, either there exists open set $U\in \mathcal{N}_x$ such that $y\not\in U$ or there exists an open set $V\in \mathcal{N}_y$ such that $x\not\in V$. In the first case (since the condition $y\not\in U$ is equivalent to saying that $U\not\in\mathcal{N}_y$) we conclude that there exists $U\in \mathcal{N}_x$ such that $U\not\in \mathcal{N}_y$. Hence $\mathcal{N}_x\ne\mathcal{N}_y$, implying that $\mathcal{N}_x\ne\mathcal{N}_y\lor \mathcal{N}_y\ne\mathcal{N}_x$. The second case can be dealt similarly.
The proof of the converse is trivial and hence skipped.

$T_1$-Space

Theorem 2. A topological space $(X,\tau)$ is $T_1$ iff $$\forall x\forall y((\{x\}\not\subseteq \{y\}\land \{y\}\not\subseteq \{x\})\iff(\mathcal{N}_x\not\subseteq\mathcal{N}_y\land \mathcal{N}_y\not\subseteq\mathcal{N}_x))$$
Proof. Let $x\ne y$ and let $X$ be $T_1$. Then since $X$ is $T_1$ by the definition of a $T_1$-space, there exists open set $U\in \mathcal{N}_x$ such that $y\not\in U$ and there exists an open set $V\in \mathcal{N}_y$ such that $x\not\in V$. Just like the proof of the previous theorem, we are then forced to conclude that there exists $U\in \mathcal{N}_x$ such that $U\not\in \mathcal{N}_y$ and there exists an open set $V\in \mathcal{N}_y$ such that $V\not\in \mathcal{N}_x$. Consequently $\mathcal{N}_x\not \subseteq \mathcal{N}_y\land \mathcal{N}_y\not\subseteq\mathcal{N}_x$. 
The proof of the converse is trivial and hence skipped.

$T_2$-Space

Theorem 3. A topological space $(X,\tau)$ is $T_2$ iff $$\forall x\forall y((\{x\}\not\subseteq \{y\}\land \{y\}\not\subseteq \{x\})\iff(\mathcal{C}_x\not\subseteq\mathcal{C}_y\land \mathcal{C}_y\not\subseteq\mathcal{C}_x))$$
Proof. Suppose $x,y\in X$ such that $x\ne y$ and $X$ is $T_2$. Since $X$ is $T_2$, there exists open sets $U, V$ such that,
  
  
*
  
*$x\in U$
  
*$y\in V$
  
*$U\subseteq X\setminus V$   
The last condition implies that $\overline{U}\subseteq X\setminus V$. But observe that $\overline{U}\not\in \mathcal{C}_y$ (because $y\not\in X\setminus V$) but $\overline{U}\in \mathcal{C}_x$. Consequently it follows that $\mathcal{C}_x\not\subseteq \mathcal{C}_y$. In a similar manner we will be able to prove that $\mathcal{C}_y\not\subseteq \mathcal{C}_x$ and this would imply in turn that $\mathcal{C}_x\not\subseteq\mathcal{C}_y\land \mathcal{C}_y\not\subseteq\mathcal{C}_x$ and we are done.
The proof of the converse is trivial and hence skipped.

$T_3$-Space

Theorem 4. A topological space $(X,\tau)$ is $T_3$ iff (here $A$ denotes a closed set) $$\forall x\forall A((\{x\}\not\subseteq A\land A\not\subseteq \{x\})\iff(\mathcal{C}_x\not\subseteq\mathcal{C}_A\land \mathcal{C}_A\not\subseteq\mathcal{C}_x))$$
Proof. Suppose $x\in X$ and $A\subseteq X$ such that $x\not\in A$ and $X$ is $T_3$. Since $X$ is $T_3$, there exists open sets $U, V$ such that,
  
  
*
  
*$x\in U$
  
*$A\subseteq V$
  
*$U\subseteq X\setminus V$   
The last condition implies that $\overline{U}\subseteq X\setminus V$. But observe that $\overline{U}\not\in \mathcal{C}_A$ (because $A\not\subseteq X\setminus V$) but $\overline{U}\in \mathcal{C}_x$. Consequently it follows that $\mathcal{C}_x\not\subseteq \mathcal{C}_A$. 
To show that $\mathcal{C}_A\not\subseteq\mathcal{C}_x$ observe that $\overline{V}\subseteq X\setminus U$. But observe that $\overline{V}\not\in \mathcal{C}_x$ (because $x\not\in X\setminus U$) but $\overline{V}\in \mathcal{C}_A$. Consequently it follows that $\mathcal{C}_A\not\subseteq \mathcal{C}_x$. This implies in turn that $\mathcal{C}_x\not\subseteq\mathcal{C}_A\land \mathcal{C}_A\not\subseteq\mathcal{C}_x$ and we are done.
To prove the converse observe that if $z\in A\cap B$ and $V\in \mathcal{C}_A$ then as $A\subseteq V$ and $x\in A$ so $x\in V$ and hence it follows immediately that $V\in \mathcal{C}_x$. Consequently it follows that $\mathcal{C}_A\subseteq \mathcal{C}_x$ and we are done.

$T_4$-Space

Theorem 5. A topological space $(X,\tau)$ is $T_4$ iff (here $A,B$ denotes closed sets) $$\forall B\forall A((B\not\subseteq A\land A\not\subseteq B)\iff(\mathcal{C}_B\not\subseteq\mathcal{C}_A\land \mathcal{C}_A\not\subseteq\mathcal{C}_B))$$
Proof. Suppose $A,B\subseteq X$ such that $A\cap B=\emptyset$ and $X$ is $T_4$. Since $X$ is $T_4$, there exists open sets $U, V$ such that,
  
  
*
  
*$B\subseteq U$
  
*$A\subseteq V$
  
*$U\subseteq X\setminus V$   
The last condition implies that $\overline{U}\subseteq X\setminus V$. But observe that $\overline{U}\not\in \mathcal{C}_A$ (because $A\not\subseteq X\setminus V$) but $\overline{U}\in \mathcal{C}_B$. Consequently it follows that $\mathcal{C}_B\not\subseteq \mathcal{C}_A$. 
To show that $\mathcal{C}_A\not\subseteq\mathcal{C}_B$ observe that $\overline{V}\subseteq X\setminus U$. But observe that $\overline{V}\not\in \mathcal{C}_B$ (because $B\not\subseteq X\setminus U$) but $\overline{V}\in \mathcal{C}_A$. Consequently it follows that $\mathcal{C}_A\not\subseteq \mathcal{C}_B$. This implies in turn that $\mathcal{C}_B\not\subseteq\mathcal{C}_A\land \mathcal{C}_A\not\subseteq\mathcal{C}_B$ and we are done.
To prove the converse without loss of generality let us assume that $B\subseteq A$. Then observe that if $B\subseteq A$ and $V\in \mathcal{C}_A$ then as $A\subseteq V$ and $B\subseteq A$ so $B\subseteq V$ and hence it follows immediately that $V\in \mathcal{C}_B$. Consequently it follows that $\mathcal{C}_A\subseteq \mathcal{C}_B$ and we are done.

$T_5$-Space (I am indebted to Daniel Wainfleet for this theorem, see his remark below)

Theorem 6. A topological space $(X,\tau)$ is $T_5$ iff $$\forall B\forall A((B\not\subseteq A\land A\not\subseteq B)\iff(\mathcal{N}_B\not\subseteq\mathcal{N}_A\land \mathcal{N}_A\not\subseteq\mathcal{N}_B))$$
Proof. Suppose $A,B\subseteq X$ and $X$ is $T_5$. Since $X$ is $T_5$, we have,
  
  
*
  
*$B\subseteq X\setminus\overline{A}$
  
*$A\subseteq X\setminus\overline{B}$
Now observe that $X\setminus \overline{A}\in \mathcal{N}_B$ and $X\setminus \overline{B}\in \mathcal{N}_A$. However, $X\setminus \overline{A}\not\in \mathcal{N}_A$ because $A\not\subseteq X\setminus \overline{A}$ (unless $A=\emptyset$, which clearly is not the case always. Consequently it follows that $\mathcal{N}_B\not\subseteq \mathcal{N}_A$. Similarly, one can prove that $\mathcal{N}_A\not\subseteq \mathcal{N}_B$  
To prove the converse without loss of generality let us assume that $B\subseteq A$. Then observe that if $B\subseteq A$ and $V\in \mathcal{N}_A$ then as $A\subseteq V$ and $B\subseteq A$ so $B\subseteq V$ and hence it follows immediately that $V\in \mathcal{N}_B$. Consequently it follows that $\mathcal{N}_A\subseteq \mathcal{N}_B$ and we are done.

Questions


*

*Are the proofs of my theorems correct?

*Is it possible to formulate equivalent conditions for other spaces mentioned here that I haven't covered only in terms of $\mathcal{N}$'s and $\mathcal{C}$'s?  
 A: You claim that $T_i$ is equivalent to an equivalence $φ \iff ψ$. But you have only proved $T_i \implies (φ \implies ψ)$ and $T_i \implies (ψ \implies φ)$. This only shows $T_i \implies (φ \iff ψ)$, the proof of the converse is missing. Also, the implication $ψ \implies φ$ is often just trivial and holds even without the $T_i$ assumption.
As @bof showed, for Hausdorff case the characteristic does not work. This is because you defined $\mathcal{C}_x$ to be the collection of all closed sets containing $x$. A better choice would be to define $\mathcal{C}_x$ as the collection of all closed neigborhoods of $x$. (Which is how you have independently edited the definition anyway.) But it turns out that the corresponding characteristic still does not work.
Consider the following properties:


*

*For every $x ≠ y ∈ X$ there is a closed neighborhood $C$ of $x$ that does not contain $y$.

*for every $x ≠ y ∈ X$ there is a closed neighborhood $C$ of $x$ that is a not a closed neighborhood of $y$.


By the motivating answer, 1. is equivalent to Hausdorff, but the characterization of your form for the modified $\mathcal{C}_x$ corresponds to 2. Clearly, this property is weaker that $T_2$ and stronger that $T_1$. The problem is that $C$ might contain $y$, but not in the interior.
In fact, one may construct a space that satisfies 2. but not 1. We take $ℕ$ and two extra points $∞_2$, $∞_3$. $\mathbb{N}$ will be open discrete and $\{ ∞_2, ∞_3\}$ will be closed discrete, so the topology will be described by the neighborhood filters for the infinities. $\mathcal{N}_{∞_2}$ will be generated by the sets $2^n ℕ$, and $\mathcal{N}_{∞_3}$ will be generated by the sets $3^n ℕ$. Note that the filters contain all cofinite subsets, so the topology is $T_1$. It is also $T_2$ for all pairs but $∞_2, ∞_3$, for which $T_2$ fails since $2^n ℕ ∩ 3^m ℕ ≠ ∅$ for every $m, n$. On the other hand, it satisfies our property 2. $\overline{2 ℕ} = 2 ℕ ∪ \{∞_2, ∞_3\}$, and it is not a neighborhood of $∞_3$ since no set $3^n ℕ$ is contained in it.
Also note that the reals with two zeros, a classical example of $T_1$ non-Hausdorff space also fails to satisfy 2.
Another observation: the condition 2. is equivalent to the following: for every $x ≠ y$ there exists a regular open set $U$ such that $x ∈ U ∌ y$. Therefore, it is equivalent to the semiregularization of the space being $T_1$. So, for semiregular spaces it is equivalent to $T_1$.
A: Your first two theorems are trivially true. The rest are blatantly incorrect. I will show this for Theorem 3.
According to your Theorem 3, a topological space $(X,\tau)$ is $T_2$ iff it satisfies the condition
$$\forall x\forall y((\{x\}\not\subseteq \{y\}\land \{y\}\not\subseteq \{x\})\iff(\mathcal{C}_x\not\subseteq\mathcal{C}_y\land \mathcal{C}_y\not\subseteq\mathcal{C}_x))\tag1$$
This is wrong, because in fact every $T_1$-space satisfies $(1)$.
Suppose $(X,\tau)$ is a $T_1$-space. I will show that $(1)$ holds.
If $(\mathcal C_x\not\subseteq\mathcal C_y\land\mathcal C_y\not\subseteq\mathcal C_x)$ then $x\ne y,$ whence $\{x\}\not\subseteq\{y\}$ and $\{y\}\not\subseteq\{x\}.$
If $(\{x\}\not\subseteq\{y\}\land\{y\}\not\subseteq\{x\}),$ then $x\ne y.$ Since $(X,\tau)$ is a $T_1$-space, $\{x\}$ and $\{y\}$ are closed sets. Hence $\{x\}\in\mathcal C_x\setminus\mathcal C_y$ and $\{y\}\in\mathcal C_y\setminus C_x,$ showing that $\mathcal C_x\not\subseteq C_y$ and $\mathcal C_y\not\subseteq\mathcal C_x.$
