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 $Z$.
First observe that $S$ is closed. It follows from the fact that if $x^*_n\in S$ and $x^*_n\to y^*$ in the weak* topology then $||y^* ||\leq R$ and $|y^* (x_i ) |\leq \alpha_i $ for all $i\in I.$ So $S$ is closed in weak* topology in $X^*.$
Moreover $S$ is convex this follows from the fact that the sets $$H^{x_i}_{\alpha_i } =\{ x^* \in X^* : |x^* (x_i )|\leq \alpha_i \}$$ are convex and $$S=\{x^*\in X:||x^* ||\leq R\}\cap\bigcap_{i\in I} H^{x_i}_{\alpha_i }$$
Moreover from the above follows that the set $S$ is also bounded.
Now observe that $$\{x^* (x_0 ): x\in S \} = k_{x_0} (S)$$ where $k_{x_0 }: X^{*} \to\mathbb{R} $ is defined by $$k_{x_0 } (x^* ) =x^* (x_0 ).$$
Moreover $k_{x_0} $ is continous when $X^*$ is equiped with weak* topology this follows from fact topology $$k_{x_0 } (\{x\in X^* : |x^* (x_0 )|<\varepsilon \} ) \subset (-\varepsilon , \varepsilon ).$$
So we prove that $S$ is closed bounded and convex set in weak* topology and hence by Banach - Alaoglu theorem it is compact in weak* topology.
The set $\{x^* (x_0 ): x\in S \} = k_{x_0} (S)$ is a image of compact convex set by a continous linear functional and hence it is compact convex subset of real line thus it is a closed interval.
Best Answer
The proof is mostly correct, except the step where you note that $\hat x\in X^{**}$: this makes it seem like you are trying to show that the weak topology is completely Hausdorff on a dual space.
That $\hat x\in X^{**}$ is immaterial, what matters here is that $x\in X$.
A different way to see that weak$^{*}$ topology is completely Hausdorff (and actually Tychonoff) - and as a consequence, so is the weak topology on any dual space, because it is finer) is to note that the weak$^{*}$ topology is just the topology of pointwise convergence on $X$.
Since complete regularity is preserved by arbitrary products and by taking a subspace, it follows that any subspace of ${\bf R}^X$ (or ${\bf C}^X$), including $X^*$, is completely regular (Hausdorff).