Connected and functionally Hausdorff with at least two points implies cardinality at least that of the continuum

Statement
Any connected functionally Hausdorff space having at least two points is uncountable. In fact, its cardinality is at least equal to the cardinality of the continuum.

Functionally Huasdorff space
A topological space $$X$$ is termed a functionally Huasdorff space if, given any two points $$x,y \in X$$, there is a continuous function $$f:X \to [0,1]$$ such that $$f(x) = 0$$ and $$f(y) = 1$$.

Related facts

 * Connected and normal Hausdorff with at least two points implies cardinality at least that of the continuum: This follows because, by Urysohn's lemma, normal Hausdorff spaces are functionally Hausdorff.
 * Connected and regular Hausdorff implies uncountable: Interestingly, this proof does not yield that the cardinality must be at least that of the continuum, only that it must be uncountable.

Proof
Suppose $$X$$ is a connected functionally Hausdorff space with at least two points. Say, $$x \ne y \in X$$ are two points. Then, by the functionally Hausdorff condition, there exists a function $$f:X \to [0,1]$$ such that $$f(x) = 0$$ and $$f(y) = 1$$.

Now, we claim that $$f$$ is surjective. Suppose not; suppose there exists $$a \in (0,1)$$ such that $$f^{-1}(a)$$ is empty. Then, $$f^{-1}[0,a)$$ and $$f^{-1}(a,1]$$ are disjoint open subsets whose union is $$X$$, and both are nonempty (because $$f(x) = 0$$ and $$f(y) = 1$$. This contradicts the assumption that $$X$$ is connected, hence $$f$$ must be surjective.

Thus, the cardinality of $$X$$ must be at least that of the continuum.