Cofinite —> Discrete adding infinite subsets what is the min number of infinite sized subsets you need to add to a cofinite topology (making it no longer cofinite) to yield the discrete topology given the underlying set is countably infinite? 
 A: Write the underlying set $X$ as the disjoint union of two countable sets $X_1$ and $X_2$. (In a bijection with $\Bbb N$, use the image of tje even and odd numbers, e.g.)
For every $x\in X$ add $X'_1=X_1\cup \{x\}$ and $X'_2 = X_2 \cup \{x\}$ to the cofinite topology. If we then generate the topology, we get $\{x\}$ as their intersection.
And once all singletons are in a generating set, we get all subsets by unions.
So we only need to add countably many generating infinite sets (2 per element).
And if we add only finitely many sets $\mathcal{F}$ to the cofinite topology $\mathcal{T}$, the resulting generated topology is compact, by Alexander's subbase lemma:
Let $\mathcal{U}$ be an open cover from the subbase $\mathcal{S}=\mathcal{T} \cup \mathcal{F}$. If we have some $O \in \mathcal{U} \cap \mathcal{T}$ then $X\setminus O$ is finite and we can cover all remaining points using finitely many elements from $\mathcal{U}$. So in the remaining case, $\mathcal{U} \subseteq \mathcal{F}$ which was already finite, so we always have a finite subcover for any subbasic cover, hence $X$ in the  generated topology by $\mathcal{S}$ is compact, so can never be the discrete topology on an infinite set, which is not compact.
So countably many are necessary and suffice.
A: Let $\ X\ $ be an infinite countable set. The OP's question is truly and simply about generating the discrete topology by a minimal set of infinite subsets of $\ X.\ $ Indeed, to obtain the discrete topology one must obviously add infinitely many sets to the topology of cofinite sets to generate all singletons. 
Actually, we may assume that
$$ X:=\mathbb Z\times\{-1\ \ 1\} $$
THEOREM   There exists a countable family $\ \mathbf B\ $ in
$\ X\ $ which consists exclusively of infinite sets and which
generates the discrete topology.
PROOF   Let $\ e=\pm 1\ $ and
$$ B_{n\ e}\ :=
  \ \big\{(k\ f)\in X:\ \big(\frac 12+k-n\big)\cdot e\cdot f > 0\big\} $$
and
$$ \mathbf B\ :=\ \{B_x:\ x\in X\} $$
Then
$$ B_{m\ d}\cap B_{n\ e}\ =
   \ \big\{(k\ f)\in X:\ \big(\frac 12+k-m\big)\cdot d\cdot f > 0
      < \big(\frac 12+k-n\big)\cdot e\cdot f \big\}  $$
hence
$$ B_{m\ -\!1}\cap B_{n\ 1}\ =
    \ \big\{(k\ f)\in X:\ \big(m-k-\frac 12\big)\cdot f\ >\ 0
  \ < \big(\frac 12+k-n)\cdot f \big\}  $$
e.g. $\ B_{n\,\ -\!1}\cap B_{n\ 1}\ =\ \emptyset\ $ but that's extra (for
fun only). What counts is that
$$ B_{n+1\,\ -\!1}\cap B_{n\ 1}\ =
    \ \big\{(k\ f)\in X:\ \big(n+\frac12-k\big)\cdot f\ >\ 0
  \ < \big(k+\frac 12-n\big)\cdot f \big\}  $$
i.e.
$$ B_{n+1\,\ -\!1}\cap B_{n\ 1}\ =\ \{(n\ 1)\} $$
is a singleton, and
$$ B_{m\ -\!1}\cap B_{m+1\,\ 1}\ =
    \ \big\{(k\ f)\in X:\ \big(m-\frac 12-k\big)\cdot f\ >\ 0
  \ < \big(k-\frac 12-m)\cdot f \big\}  $$
i.e.
$$ B_{m\ -\!1}\cap B_{m+1\,\ 1}\ =\ \{(m\,\ -\!1)\} $$
is a singleton again.
Thus, with respect to the topology generated by $\ \mathbf B,\ $
every singleton $\ \{x\}\ $ in $\ X\ $ is isolated, i.e. the topology
is discrete.   End of Proof
A: 
There was still the issue of the minimality of the addition to the starting topology. The following theorem will address this in full generality.

THEOREM Let $\ (X\ T)\ $ be an arbitrary topological space. Let
$\ \mathbf B\ $ be a family which generates the discrete topology in $\ X.\ $
Then cardinality $\ |\mathbf G\setminus T|\ $ is at least as large as
the cardinality of the set of the non-isolated points of $\ (X\ T).$
PROOF   Let $\ x\in X\ $ be non-isolated. Then there exists
$\ G_x\in\mathbf B\ $ such that $\ x\ $ is the only non-isolated
point in $\ G_x.\ $ Obviously, $\ G_x\not\in T,\ $ and equally obviously
$\ G_x\cap G_y=\emptyset\ $ whenever $\ x\ne y,\ $ i.e. for different
non-isolated points we get different members of $\ \mathbf B\setminus T.\ $
  End of Proof
