Yes, there is such a space. Let $X=2^{\omega_1}$ be the space of binary sequences of length $\omega_1$, in the order topology generated by the lexical order. So $X$ consists of the branches through the tree $2^{<\omega_1}$, with the left-to-right order on branches. This is an order topology of a linear order and hence Hausdorff.
The key thing to notice is that every element $a\in X$ is the limit of an $\omega_1$-sequence. If $a_\alpha\to a$ for $\alpha<\omega_1$ and $f:X\to\mathbb{R}$ is continuous, it follows that $f(a_\alpha)\to f(a)$. Since every convergent $\omega_1$-sequence in the reals is eventually constant, it must be that $f(a_\alpha)=f(a)$ for all sufficiently large countable ordinals $\alpha$. So this space has your desired property.