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?
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:
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 $T_1$.