Is the fundamental group of a non-semi-locally simply connected space uncountable?

Question

I know that the fundamental group of the Hawaiian earring is uncountable, so I would think you can argue it injects into such a space, but I think that would rely on some condition like having a countable neighborhood basis.

Answer 1

Here is a proposed counterexample to my own question:

Let $P$ be an ordered set with such that there is no order preserving function $f$ from the natural numbers that has the property that for any element $x \in P$ there is an integer $y$ such that $x \leq f(y)$.

Put a topology on $X=P\cup \{*\}$ by declaring as a basis the individual points in $P$ and the sets containing $*$ such that there is an element $x \in P$ in the set and that if $x \leq y$, then $y$ is in the set. If $P$ were the natural numbers this would be the set $\{0\}\cup\{1/n | n \in \mathbb{N}\}$ with the subspace topology.

Take the reduced suspension of $X$ where the basepoint is $*$ and attach a disk from the arc corresponding to the element $x \in P$ to the arc corresponding to $y \in P$ if $x < y$. Call this space $X'$.

I claim $X'$ is not semi-locally simply connected. Any neighborhood around the basepoint contains some circle corresponding to some element $x \in P$. It is pretty clear that the loop obtained by traversing this circle once can only be contracted by moving toward the basepoint through these arcs we added. However, a homotopy can only take us through countably many of these arcs so by the condition on $P$ this homotopy cannot end at the basepoint, so the loop is not trivial.

The fundamental group is also not uncountable because it's pretty clear the loops corresponding to the elements of $P$ have to be the building blocks, but there cannot be infinite sequences of them like in the Hawaiian earring because by the condition on $P$ these loops are not approaching the basepoint. In fact, I think that the fundamental group is $\mathbb{Z}$ generated by the loop corresponding to some element of $P$.

Again this is only a sketch, please dissect the argument because I would like it to be the case the titular question has an affirmative answer.