Questions about existence of injections between infinite sets and the sets of all infinite topologies on them 1) If $X$ is an infinite set and $T_X$ the set of all infinite topologies on $X$ is it in general true that there is no injection $f_T:T_X \to X$?
2) What conditions on $X$ assure an injection (if that´s ever possible)?
By infinite topologies, I mean topologies with an infinite number of sets (subsets of $X$) as its elements.
 A: Similar to Keith Kearnes’ answer, but made slightly more concrete:


*

*there’s an injection from “partial orders on $X$” to “infinite topologies on $X$”, sending any partial order to its Alexandrov topology, whose open sets are just the down-closed sets of the order.  This is always infinite if $X$ is, since the sets $\mathord{\downarrow} \{x \}$ are distinct for each $x \in X$; and it’s an injection since any partial order can be recovered from its Alexandrov topology as the specialisation order.

*there’s an injection from “proper subsets of $X$” to “partial orders on $X$” sending $S \subseteq X$ to the order where $x <_S y$ just if $x \in S$ and $y \notin S$.  (I.e. all of $X \setminus S$ is above all of $S$, and the order is otherwise discrete.)  This is an injection since $S$ can be recovered as $\{ x \mathrel{|} \exists y,\ x <_S y \}$.

*composing these gives an injection from “proper subsets of $X$” to “infinite topologies on $X$”

*so any injection “infinite topologies on $X$” to $X$ would give an injection from “proper subsets of $X$” to $X$, which is impossible.
I’m assuming throughout that you work in ZFC.  Without choice, the earlier parts all work, but the final step breaks, because we don’t know that there’s no injection $\mathcal{P}(X) \to X$.
In a comment, you mention restricting further, to topologies whose (nonempty) open sets are all infinite.  That doesn’t essentially change anything, as can be shown by a slight modification of the above construction.  Fix a countable and co-infinite set $K \subseteq X$.  Now take the injection from “proper subsets of $X \setminus K$” to preorders on $X$, where $\leq_S$ puts $K$ codiscretely at the bottom, $S$ discretely in the middle, and $X \setminus (S \sup K)$ discretely on top.  Now send this to its specialisation order.  This gives an injection from “proper subsets of $X \setminus K$” to “infinite topologies on $X$ with all nonempty opens infinite”.  But since $K$ was countable and co-infinite, $|X \setminus K| = |X|$, so we are done as before.
A: Let
$$ \forall_{A\subseteq X}\quad \mathcal B_A
\ :=\ \{\{a\} :\ a\in A\}\cup X $$
Then each $\ \mathcal B_A\ $ is a topological base in $\ X\ $
while every two of them generate different topologies in $\ X.\ $ When $\ X\ $ is infinite then
$$ |\,\{\mathcal B_A :
    \, A\subset X\ \ \text{and}\ \ |A|=\infty \}\,|
      \,\ =\,\ 2^{|X|} $$ 
We see that an injection of the set of all topologies
in $\ X\ $ into $\ X\ $ is impossible when $\ X\ $ infinite.
Remark 1    There is no such injection also when $\ X\ $ is finite but $\ |X|\ne 1.\ $
Indeed, let
$$ \forall_{A\subseteq X}\quad T_A
           \ :=\ \{\emptyset\,\ A\,\ X\} $$
These are topologies, and
$$ |\,\{T_A:\,A\subseteq X\}\,|\ =\ 2^{|X|}-1 $$
for every non-empty finite $\, X.\, $ In the empty case we get $\, 0\, $ points and $\, 1\, $ topology, $\ 0<1.$
Remark 2   The situation, cardinality wise, is even more dramatic.
A: For an infinite set $X$, there are $2^{2^{|X|}}$ topologies with infinitely many elements on $X$.
The upper bound is obvious, the lower bound is given by my comment (and the slight correction in YCor's): for any ultrafilter $U$ on $X$, define $T_U$ to have as opens the empty set and the elements of $U$. It is clear that this forms a topology, and that you can recover $U$ from it. 
Moreover, clearly $T_U$ has infinitely many open sets. 
So $U\mapsto T_U$ defines an injection $\beta X\to T_X$, and it's well known that $|\beta X| = 2^{2^{|X|}}$.
So there can be no injection $T_X\to X$ (in fact, no injection $T_X\to P(X)$)
YCor's comment below mine elaborates a bit on what one can ask of these topologies and how many we get (Hausdorff, metrizable)
A: If $X$ is an infinite set, then there are $2^{|X|}$-many ways to well-order $X$. Each such order determines an Alexandroff topology on $X$, which will have infinitely many open sets. The order is recoverable from the topology, so there are too many different infinite topologies on $X$ for there to be an injection $f_T: T_X\to X$.
