Showing every connected regular space having more than one point is uncountable without using proof by contradiction The common proof goes like this:

Suppose $X$ is countable, then it must be Lindelöf. A regular Lindelöf space is normal (akin to the proof that a regular and second-countable space is normal). Using Urysohn's Lemma, it's easy to establish a contradiction.

Out of curiosity, does this result have a direct proof (without assuming $X$ is countable or making similar assumptions at the beginning)? It seems it's crucial here, otherwise, we couldn't deduce $X$ is normal, hence the proof couldn't follow. I've done some research on the internet:


*

*Show that a connected regular space having more than one point is uncountable

*Is every connected regular space having more than one point uncountable?

*https://topospaces.subwiki.org/wiki/Connected_regular_space
All are of the same proof. Can we avoid proof by contradiction? Thanks in advance.
 A: The conclusion of the theorem is that $X$ is uncountable. This is defined, usually, as "$X$ is not countable". And logically speaking (in first order logic) the way to prove a statement of the form $\lnot \phi$ is to assume $\phi$ and derive a contradiction from it (even Brouwer, in his intuitionistic way of doing maths, did it that way; he did not accept, though, that proving $\lnot (\lnot \phi)$ proved $\phi$, as in classical logic). So the going via contradiction seems quite unavoidable. 
Maybe you could define "uncountable" in a positive way (e.g. there is an injection from $\aleph_1$ into $X$), but then the proof of equivalence with the "negative" notion would (I think, correct me if I'm wrong) require a form of the Axiom of Choice (another source of possible controversy). And I don't see a proof route there, personnally, so that wouldn't solve it either.
So I say, embrace it, or state the whole result "positively", in a way that does not require a proof by contradiction:

Let $X$ be a countable, regular ($T_3$) space with $|X|>1$. Then $X$ is disconnected.

Proof: $X$ is Lindelöf, so $X$ is $T_4$. Pick $x \neq y$ in $X$ and let $f: X \to [0,1]$ be continuous with $f(x)=0, f(y)=1$ (Urysohn lemma applied to $\{x\}$ and $\{y\}$). Then $f[X]$ is countable and $[0,1]$ is not, so pick $t \in [0,1]\setminus f[X]$ and then $f^{-1}[[0,t)]$ and $f^{-1}[(t,1]]$ form a decomposition of $X$. So $X$ is disconnected.
