"big" Hausdorff space with dense subspace of given cardinality In a topology course we proved the following proposition:

Let $A$ be an infinite set. Then there exists a Hausdorff space $X$ of cardinality $|\mathfrak{P}(\mathfrak{P}(A))|$ which contains a dense subspace of cardinality $|A|$.

This proposition has nice consequences - it shows e.g., that there are $|\mathfrak{P}(\mathfrak{P}(A))|$-many (ultra-)filters on any infinite set $A$. However, the only proof I know is horrifically technical. It takes the product topology on $X := A^{\mathfrak{P}(A)}$ (where $A$ is considered to be discrete) and constructs a complicated subspace for which it is not trivial to see, that it is of cardinality $|A|$.
I am looking for an elegant proof which is more easy to understand. It is not important for me, whether it uses the same construction or another one, but another construction would of course be particularly interesting :)
Thank you in advance!
 A: The Added part of this answer contains a proof that there are $2^{2^\kappa}$ ultrafilters on $\kappa$ and hence that $|\beta D|=2^{2^\kappa}$ if $D$ is the discrete space of cardinality $\kappa$. (It wouldn’t surprise me if the construction that you mention in the question is actually a disguised form of the same idea, or at least something similar.)
A: No filters here.
$(1).$ Notation:
$|X|$ is the cardinal of a set, or space, $X.$
$d(U)$ ( the density of $U$ ) is the least infinite cardinal $k$ such  that the space $U$ has a dense subset $V$ with $|V|\leq k.$
$w(U)$  ( the weight of $U$ ) is the least infinite cardinal $k$ such that the space $U$ has a base (basis) $B$ with $|B|\leq  k$.
$D(m)$ is "the" discrete space  with $|D(m)|=m.$
"Product space" topology means  the Tychonoff product-topology.
$(2).$ The Hewitt-Marczewski-Pondiczery Theorem (H-M-P): Let $m$ be an infinite cardinal. If  $d(X_s) \leq m$  for each $s\in S,$ and if  $|S|\leq 2^m$  then $d(\; \prod_{s\in S}X_s\;)\leq m.$
In particular suppose each $X_s=D(m)$  and  $|S|=2^m.$  A product of Hausdorff spaces is Hausdorff so the space $P=\prod_{s\in S}X_s$ is Hausdorff. We have $|P|=|D(m)^{2^m}|=|m^{2^m}|=2^{2^m}$ and by the H-M-P theorem we have $d(P)=m.$
In "General Topology" by R. Engelking,  this is Theorem 2.3.15 (i.e. Chapter 2, Section 3, Proposition 15).
If you want to read it without reading the previous 80 pages, this may help: The previous results that he quotes are:
(a)... 2.3.5. If each $X_s\ne \emptyset$ and $A_s \subset X_s$ then $\prod_{s\in S}A_s$ is dense in $\prod_{s\in S}X_s$ iff  each $A_s$ is dense in $X_s$    
(b)... 1.4.10. If $f:X\to Y$ is a continuous surjection then $d(Y)\leq d(X).$ 
(c)... From  the def'n of the Tychonoff product  it is easily seen that $w(D(m)^m)=m$ ( which  is used in his proof of the H-M-P theorem, where  he refers to $D(m)^m$ as $T$). 
Note: (a) and (b) are only used in the first part of the proof, to show that it suffices to prove that $d(D(m)^{2^m})=m.$ 
