Is there a Hausdorff space $X$ separating points of any Hausdorff space?

There does not exist such $X$.

Given any Hausdorff space $X$, there exists a Hausdorff $Y$ and distinct $a,b\in Y$ such that each continuous $f\colon Y\to X$ satisfies $f(a)=f(b)$.

This follows from the following more precise statement, which I will prove.

Let $X$ be Hausdorff with cardinality $\kappa=\lvert X\rvert$, and $\kappa^\prime$ be a cardinal such that there is no finite-to-one mapping $\kappa^\prime\to\kappa$. Then, there exists a Hausdorff space $Y$ with distinct points $a,b\in Y$ and subset $Z\subseteq Y$ satisfying

1. Every closed neighborhood of $a$ or $b$ contains all but finitely many points of $Z$.
2. $Z$ has cardinality $\kappa^\prime$.

Furthermore, for any such $Y$, every continuous $f\colon Y\to X$ satisfies $f(a)=f(b)$.

The statement that there is no finite-to-one mapping $\kappa^\prime\to\kappa$ just means that there is no function $g\colon\kappa^\prime\to\kappa$ with finite inverse images $g^{-1}(k)$ for all $k\in\kappa$. It can be seen that $\kappa^\prime=\aleph_0^\kappa$ works.

I'll first show that if $Y$ is chosen as above then all continuous $f\colon Y\to X$ satisfy $f(a)=f(b)$. Then, I'll show how to construct $Y$.

So, suppose that $f\colon Y\to X$ is continuous with $f(a)\not=f(b)$. As $X$ is Hausdorff, there are disjoint open neighbourhoods $U,V$ of $f(a),f(b)$. Also, as $Z$ has cardinality $\kappa^\prime$, there exists a point $x\in X$ with $f^{-1}(x)$ containing an infinite subset of $Z$. As the closure of $f^{-1}(V)$ contains all but finitely many points of $Z$, there exists a point $c$ in the intersection of $Z$, $f^{-1}(x)$ and the closure of $f^{-1}(V)$. In particular, $c$ cannot be contained in $f^{-1}(U)$, otherwise $f^{-1}(U)\cap f^{-1}(V)=f^{-1}(U\cap V)$ would be nonempty. This means, in particular, that $f(a)\not= f(c)=x$. Then, as $X$ is Hausdorff, there exist disjoint open neighbourhoods $U^\prime,W$ of $f(a)$ and $x$. Then, $f^{-1}(U^\prime)$ is an open set containing $a$ whose closure is contained in $Y\setminus f^{-1}(W)\subseteq Y\setminus f^{-1}(x)$, contradicting the fact (from 1) that it contains all but finitely many points of $Z$. QED

Now, I'll construct the Hausdorff space $Y$. Let the set of points in $Y$ be the collection of the following distinct points.

• $a$ and $b$.
• $a_{k,n}$ and $b_{k,n}$ for $k\in\kappa^\prime$ and $n\in\mathbb{N}$.
• $c_k$ for $k\in\kappa^\prime$.

Also, let $Z=\lbrace c_k\colon k\in\kappa^\prime\rbrace$, which has cardinality $\kappa^\prime$. Let $\mathcal U$ be the collection of subsets of $Y$ of the following form.

1. $\lbrace a\rbrace\cup\lbrace a_{k,n}\colon k\in\lambda,n\in\mathbb{N}\rbrace$ for any cofinite subset $\lambda$ of $\kappa^\prime$.
2. $\lbrace b\rbrace\cup\lbrace b_{k,n}\colon k\in\lambda,n\in\mathbb{N}\rbrace$ for any cofinite subset $\lambda$ of $\kappa^\prime$.
3. $\lbrace a_{k,n}\rbrace$ for any $k\in\kappa^\prime$ and $n\in\mathbb{N}$.
4. $\lbrace b_{k,n}\rbrace$ for any $k\in\kappa^\prime$ and $n\in\mathbb{N}$.
5. $\lbrace c_k\rbrace\cup\lbrace a_{k,n}\colon n\ge N\rbrace\cup\lbrace b_{k,n}\colon n\ge N\rbrace$ for any $k\in\kappa^\prime$ and $N\in\mathbb{N}$.

It can be seen that the intersection of any two elements of $\mathcal{U}$ is a union of elements of $\mathcal{U}$, so it defines a topology for $Y$, which it can be checked is Hausdorff. Finally if $U$ is a neighbourhood of $a$ or $b$, then it contains a set of the form (1) or (2) for some cofinite subset $\lambda$ of $\kappa^\prime$ and, hence, its closure contains the cofinite subset $\lbrace c_k\colon k\in\lambda\rbrace$ of $Z$.