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
- Every closed neighborhood of $a$ or $b$ contains all but finitely many points of $Z$.
- $Z$ 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.
- $\lbrace a\rbrace\cup\lbrace a_{k,n}\colon k\in\lambda,n\in\mathbb{N}\rbrace$ for any cofinite subset $\lambda$ of $\kappa^\prime$.
- $\lbrace b\rbrace\cup\lbrace b_{k,n}\colon k\in\lambda,n\in\mathbb{N}\rbrace$ for any cofinite subset $\lambda$ of $\kappa^\prime$.
- $\lbrace a_{k,n}\rbrace$ for any $k\in\kappa^\prime$ and $n\in\mathbb{N}$.
- $\lbrace b_{k,n}\rbrace$ for any $k\in\kappa^\prime$ and $n\in\mathbb{N}$.
- $\lbrace c_k\rbrace\cup\lbrace a_{k,n}\colon n\ge N\rbrace\cup\lbrace b_{k,n}\colon n\ge N\rbrace$ 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 $Z$.
Take Y as in the remark of the article by A.Mysior you have linked. I will follow its notations.
As it is said in the article, Y is regular.
Now take any real valued continuous function $f$ on Y.
For $\epsilon >0$, $\exists N \in \mathbb{N} $ such that $U_N(b) \subset f^{-1}([f(b)- \epsilon , f(b) + \epsilon])$ (because since f is continuous, $f^{-1}([f(b)- \epsilon , f(b) + \epsilon])$ is a neighborhood of $b$) .
Set $J_n = f^{-1}([f(b)- \epsilon , f(b) + \epsilon]) \cap \{(x,0) \ | \ n-1\leq x \leq n \}$, since $J_{-N-1}$ is infinite, you can apply the same inductive reasoning as in the article (where it is done for $K_n$), hence $J_n$ is infinite $\forall n \in \mathbb{Z}$.
By continuity, we get $a \in f^{-1}([f(b)- \epsilon , f(b) + \epsilon])$.
Thus $f(a)=f(b)$.
I hope this can help.
Best Answer
Yes, $X$ is completely Hausdorff. There are two nontrivial cases: separating two points of $\mathbb{R}\times\{0\}$ and separating a point of $\mathbb{R}\times\{0\}$ and $a$. For both of these cases, observe that for any $x\in\mathbb{R}$ the set $U_x=\{(x,0)\}\cup I_x\cup I_x'$ is clopen in $X$. Thus the characteristic function of $U_x$ is continuous and separates $(x,0)$ from every other point of $\mathbb{R}\times\{0\}$ and from $a$.