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.