Given $X=\{1,\tfrac{1}{2}, \tfrac{1}{3}, \dots\} \cup \{0\}$, why is $\pi_1(SX)$ countable while $\pi_1(\Sigma X)$ is uncountable? Letting $X=\{1,\tfrac{1}{2}, \tfrac{1}{3}, \dots \}\cup \{0\}$, $SX$ the suspension, and $\Sigma X$ the reduced suspension after contracting the segments connected to 0 to a point. I know that $\Sigma X$ is equivalent to the Hawaiian earring. I'm aware of (what I have been lead to believe is) a proof that its fundamental group is uncountable: 

Given any subset $\{n_1, n_2, \dots\}\subset \mathbb{N}$, we can create a loop that travels around the circle of radius $\tfrac{1}{n_1}$, followed by the circle of radius $\tfrac{1}{n_2}$, and so on, meaning loops in $\Sigma X$ are in bijection with $2^{\mathbb{N}}$ and therefore uncountable.

My question is as follows: why doesn't this hold for $SX$ which, according to Hatcher Exercise 1.2.18, is countable. Starting from the top point, couldn't I create uncountably many distinct loops in a similar manner? What is it about collapsing the segments attached at zero that creates an uncountable group?
 A: The same construction doesn't work for $SX$.
Note that the sequence of paths is typically combined into a single path by putting each path on a subinterval of $[0,1]$ with $1$ being a limit of those subintervals. Now with $\Sigma X$ we can map $1$ to the unique shared point. So every sequence covergent to $1$ will be mapped to a sequence converging to the unique point. But with $SX$ we can find a sequence converging to $1$ but its image converges to any point lying at the vertical line over $0$. There is no valid choice for value at $1$ making the construction continuous.
This also shows that any loop in $SX$ can go around only finitely many distinct subcircles of $SX$. You can conclude from that that the fundamental group is countable.

Let's dive into details. Let $SX=(X\times [0,1])/\sim$ and let $v_0=[(0,1)]_\sim$ be the top vertex. By $k$'th line I will understand the image of $\{1/k\}\times[0,1]$ in $SX$ and denote it by $L_k$. Note that $L_0$ will be the image of $\{0\}\times[0,1]$.
Your construction is as follows: for any sequence of naturals $n_1,n_2,\ldots$ let $f_k:[1/k,1/(k+1)]\to SX$ be a path such that $f(1/k)=f(1/k+1)=v_0$ and such that $f$ goes through $L_{n_k}$ line and back through say fixed $L_1$ line (so that they are pairwise non-homotopic). Finally we compose all $f_k$ into $f:[0,1]\to SX$ via $f(x)=f_k(x)$ if $x\in[1/k,1/(k+1)]$ and $f(0)=v_0$.
Note that this construction is continuous over $\Sigma X$ but not over $SX$. Indeed, let $w_k=[(1/k, 1/2)]_\sim$ and note that $w_k\to [(0,1/2)]_\sim\neq v_0$. But $f^{-1}(w_{n_i})$ is a single point that belongs to some $[1/t,1/(t+1)]$. So it forms a sequence convergent to $0$. This is a contradiction since the image does not converge to image of $0$ being $v_0$.
The main difference between $SX$ and $\Sigma X$ is that $\Sigma X$ is locally connected unlike $SX$. This implies that:

Lemma. Let $f:[0,1]\to SX$ be a continuous function. Then there are at most finitely many $k$ such that $L_k\subseteq im(f)$.

Proof. Assume that's not the case, so we have $L_{m_1},L_{m_2},\ldots$ fully contained in $im(f)$. Since $im(f)$ is compact then $$\overline{\bigcup_{i=1}^\infty L_{m_i}}\subseteq im(f)$$ This implies (by the intrinsic properties of $X$) that $L_0\subseteq im(f)$. But then $im(f)$ is not locally connected. Contradiction, since $f$ is a quotient map (onto its image) from a locally connected space (see this). $\Box$
Side note: another difference is that $SX$ is not an image of any path but $\Sigma X$ is (by the Hahn-Mazurkiewicz theorem, or by the mentioned construction).

Conclusion: $\pi_1(SX)$ is countable.

Sketch of the Proof. There's a countable number of subcircles (basically a subcircle is a pair $(L_i,L_j)$ of lines) in $SX$. Since every path goes around only finitely many of them then it means that to any path we can associate a sequence $(n_1,n_2,\ldots)\in\mathbb{Z}^\infty$ of corresponding winding numbers. Only finitely many entries are non-zero. And there are only countable many such sequences. $\Box$
