Countable subspace of uncountable set is indiscrete or $T_0$ I'm solving the problems from "Topology Without Tears"

Let $(X,\mathcal{T})$ be a topological space, where $X$ is an infinite set. Prove that $(X,\mathcal{T})$ has a subspace homeomorphic to $(\mathbb{N},\mathcal{T}_1)$, where either $\mathcal{T}_1$ is the
  indiscrete topology or $(\mathbb{N},\mathcal{T}_1)$ is a $T_0$-space.

My half-proof:
$X$ can be countable or uncountable. Number of open sets in $\mathcal{T}$ can be countable or uncountable.
There are 4 cases:
a) $X$ - infinite, countable. $\mathcal{T}$ - finite
b) $X$ - infinite, countable. $\mathcal{T}$ - infinite, countable
b2) $X$ - infinite, countable. $\mathcal{T}$ - infinite, uncountable
c) $X$ - infinite, uncountable. $\mathcal{T}$ - countable
d) $X$ - infinite, uncountable. $\mathcal{T}$ - uncountable
Cases:
c) Let's consider the case when $X$ - uncountable and $\mathcal{T}$ - countable.


*

*We can build a partition $\mathcal{P}$ (not necessarily open sets) of $X$ such that for any $p \in \mathcal{P}$ and for any open set $s \in \mathcal{T}$, the intersection $p \cap s = p$ or $\emptyset$. Also number of elements of $\mathcal{P}$ is less than or equal to $\mathcal{T}$, hence $\mathcal{P}$ is countable.

*Since $X$ is uncountable (by definition) and $\mathcal{P}$ has countable number of elements, then one element $p \in \mathcal{P}$ is an uncountable subset of $X$.

*Subset $p$ is uncountable and for any open set $s \in \mathcal{T}$, $p \cap s = p$ or $\emptyset$. Then there is an infinite countable subset $n \subset p$ and subspace $(n, \mathcal{T}_n)$ with topology induced by $\mathcal{T}$ which is homeomorphic to $(\mathbb{N},\mathcal{T}_1)$. And $\mathcal{T}_n$ is indiscrete.
a) $X$ - infinite, countable. $\mathcal{T}$ - finite.
We can build a indiscrete subspace using algorithm similar to case c). If $X$ is infinite and $\mathcal{T}$ is finite. Then at least one open set is infinite and we can find a indiscrete subspace.
b2) X - infinite, countable.  - infinite, uncountable
My intuition tells me that $X$ has an infinite countable subset with discrete topology => we can find $T_0$ subspace.
I don't know how to prove b) and d). If $X$ is a $T_0$ space than it's easy to prove. The counter example is a product of R*R with usual and indiscrete topology (which is not $T_0$) - in this case if we take one point from every indiscrete R then we will get $T_0$ subspace.
Is my overall reasoning correct? May be there is an easy proof?
 A: For $x\in X$ let $\mathcal N(x)=\{U\in\mathcal T:x\in U\},$ i.e., $\mathcal N(x)$ is the set of all open neighborhoods of $x.$
Call two points $x,y\in X$ equivalent if $\mathcal N(x)=\mathcal N(y).$ This is clearly an equivalence relation. We consider three cases.
Case 1. Some equivalence class $A$ is infinite.
The induced topology on $A$ is indiscrete. Since $A$ is infinite, it has a subset $B$ which is countably infinite. The subspace $B$ is homeomorphic to $\mathbb N$ with the indiscrete topology.
Case 2. There are infinitely many equivalence classes.
We get an infinite $\text T_0$-subspace $A$ of $X$ by choosing one point from each equivalence class. A countably infinite subspace of $A$ will be homeomorphic to $\mathbb N$ with a topology making it a $\text T_0$-space.
Case 3. Each equivalence class is finite, and the number of equivalence classes is finite.
Then $X$ is finite, contradicting our assumption that $X$ is infinite.
A: Notation:  $[S]^2$ is the set of $2$-member subsets of $S,$ for any set $S.$
The closure bar will denote closure in $X.$
There is an elementary theorem in infinitary  combinatorics which set-theorists write as $$\omega\to (\omega)^2_2$$ which means that  for any infinite $X$ and any function $f:[X]^2\to \{0,1\}$ there is a countably infinite $Y\subset X$ such that $$\{f(u):u\in [Y]^2\}=\{0\} \lor \{f(u):u\in [Y]^2\}=\{1\}.$$   
For $u=\{p,q\}\in [X]^2$ let $f(u)=0$ if $(p\in \overline {\{q\}} \land q\in \overline {\{p\}}).$ Otherwise let $f(u)=1. $
Let $Y$ be a countably infinite subset  of $X$ such that $\{f(u):u\in [Y]^2\}=\{j\},$ where $j=0$ or $j=1.$  
If $j=0$ then $p\in \overline {\{q\}}$ for all $p,q\in Y$ so $Y$ is an indiscrete subspace of $X.$ 
If $j=1$ then for every $\{p,q\}\in [Y]^2$ we have $(\;p\not \in \overline {\{q\}} \lor q\not \in \overline {\{p\}}\;),$ so $Y$ is a $T_0$ subspace of $X$.
