TITLE: For any set $X$, is there a Hausdorff $Y$, and a disjoint family $\{Y_x:x\in{X}\}$ of dense subsets?
QUESTION [8 upvotes]: Greets
I want to prove the following:
For any set $X$ there exists a Hausdorff space $Y$ and a family $\{Y_x:x\in{X}\}$ of disjoint subsets of $Y$, each dense in $Y$.
I want to prove this assertion since by proving it, I can use it to prove that any topological space is the open continuous image of a Hausdorff space.
I have proven the assertion, but using mathematical logic; namely, using the compactness theorem and the fact that the assertion is true for finite $X$ in some ordered set, but I don't know how else to prove this without the use of the compactness theorem.
I would thank very much to see a prove of this, without using mathematical logic.

REPLY [3 votes]: An answer using mathematical logic is as follows:
Let $L=\{<\}\cup{\{P_x:x\in{X}\}},$ where the $P_x$ are distinct unary predicates.
For each $x\in{X}$ let $\varphi_x=\forall{y}\forall{z}(y<z\rightarrow\exists{w}(P_xw\wedge{y<w<z}))$ and for any $x,y\in{X}$ with $y\neq{x}$ let $\Delta_{x,y}=\forall{z}¬(P_xz\wedge{P_yz})$. Let $T=\{\varphi_x:x\in{X}\}\cup{\{\Delta_{x,y}:x,y\in{X}\wedge{x\neq{y}}\}},$ then $T$ is finitely satisfiable, since $(\mathbb{Q},<)$ can be partitioned in $n$ dense subsets for any $n<\omega$. Thus by the compactness theorem there exists a model $(M,<,P_x)_{x\in{X}}$ of $T$, so the $P_x^{\mathcal{M}}$ are dense disjoint subsets of $(M,<)$, but since order topologies are Hausdorff, $(M,<)$ is our required space.