TITLE: Is it true that a second-countable topological space which has an universal cover must have countable $\pi_1$?
QUESTION [6 upvotes]: Question: Is there a second-countable, connected, locally path-connected, semi-locally simply connected (and "perhaps" Hausdorff) topological space $X$ such that $\#\pi_1(X)>\aleph_0$?

Uninteresting story of the question: I was doing an exercise which asked to prove that a Hausdorff, locally compact, second-countable, connected topological manifold must have countable fundamental group.
My idea was observing that, by Poincaré-Volterra theorem, its universal cover is second-countable, so its fibres are countable.
However, the exercise hinted towards a more direct, and somehow visual, proof, which I decided to follow. Jokes on me, the work I did implies (in my opinion) way too much. In fact, just by "following the instructions", I ended up using only N2 and the "triple connection" hypothesis that guarantees the existence of the universal covering. Not even T2, actually. I wanted to be sure I was wrong, before diving back in, though.

REPLY [2 votes]: This is a job for topological fundamental groups. In fact one can weaken the hypotheses a little bit to include some non-locally path-connected spaces.
Theorem: If $X$ is second countable and admits a simply connected covering space, then $\pi_1(X,x)$ is countable.
Let $\Omega(X,x)$ be the based loop space of $X$ with the compact-open topology. Let $\pi_{1}^{qtop}(X,x)$ be the fundamental group equipped with the quotient topology with respect to the map $q:\Omega(X,x)\to\pi_{1}^{qtop}(X,x)$, $q(\alpha)=[\alpha]$ identifying homotopy classes of loops. Warning: $\pi_{1}^{qtop}(X,x)$ is a quasitopological group (in particular is homogeneous) but may not be a topological group.
Claim 1: $\pi_{1}^{qtop}(X,x)$ is separable.
Proof. Since $S^1$ is locally compact Hausdorff and $X$ is second countable, $\Omega(X,x)$ is second countable (see Engelking 3.4.16). Every second countable space is separable and every continuous image of a separable space is separable. Therefore $\pi_{1}^{qtop}(X,x)$ is separable.
Claim 2: $\pi_{1}^{qtop}(X,x)$ is discrete.
Proof. The details are a bit technical but it is known that a covering map (and even more general a semicovering map) $p:\tilde{X}\to X$ induces a homomorphism $p_{\#}:\pi_{1}^{qtop}(\tilde{X},\tilde{x})\to \pi_{1}^{qtop}(X,x)$ which is also an open embedding. If $p$ is the universal cover, then $\pi_{1}^{qtop}(\tilde{X},\tilde{x})=1$ is the trivial discrete group. Since the trivial subgroup $p_{\#}(\pi_{1}^{qtop}(\tilde{X},\tilde{x}))=1$ is open and $\pi_{1}^{qtop}(X,x)$ is homogeneous, $\pi_{1}^{qtop}(X,x)$ is a discrete group. Under your original assumptions the simpler arguments in this paper suffice to prove Claim 2.
Proof of Theorem. Every separable discrete space is countable.