Consider $S^1 \subseteq \mathbb{C}$ with the Euclidean metric. Let $X = \Pi_{n=1}^\infty S^1$ as a set and define
$d_\infty: X \times X \to [0, 2], d_\infty((x_n)_{n=1}^\infty, (y_n)_{n=1}^\infty) = \operatorname{sup}_{n=1}^\infty d(x_n, y_n)$
Now $(X, d_\infty)$ is a metric space. What is the fundamental group of this space?
Let $G = \{(a_n)_{n=1}^\infty \in \Pi_{n=1}^\infty\mathbb{Z} \mid \exists C \in \mathbb{Z}_{\geq 0} \forall n \in \mathbb{N} : |a_n| \leq C\}$. I believe that we should have $\pi_1(X) = G$. The corresponding paths are defined by $\varphi_{(a_n)_{n=1}^\infty}: S^1 \to X, z \mapsto (z^{a_n})_{n=1}^\infty$ for $(a_n)_{n=1}^\infty \in G$.
The reason I believe that $\pi_1(X) = G$ is that the fundamental group of $X$ endowed with the product topology is simply $\Pi_{n=1}^\infty\mathbb{Z}$, and $G \subseteq \Pi_{n=1}^\infty\mathbb{Z}$ is precisely the subgroup of elements whose corresponding paths are still continuous when viewed as mappings from $S^1$ to $X$ equipped with $d_\infty$.
Is this correct?
Suppose $f:S^1\to X$ is a loop. By compactness, $f$ is uniformly continuous, and it follows that the winding numbers of the coordinates of $f$ are bounded (if $\delta>0$ is such that $|x-y|<\delta$ implies $d_\infty(f(x),f(y))<1$, say, then you can bound the winding number of each coordinate of $f$ in terms of $\delta$). So there is a homomorphism $p:\pi_1(X)\to G$ which takes each loop to its sequence of winding numbers on each coordinate. As you have observed, $p$ is surjective, so it suffices to check that $p$ is also injective.
So, suppose $p([f])=0$, so $f=(f_n)$ has winding number $0$ on each coordinate. Note that each $f_n$ has a canonical nullhomotopy given by lifting it to the universal cover $\mathbb{R}$ and using the linear homotopy. It is then easy to check that these linear nullhomotopies can be joined together to give a nullhomotopy of $f$ (because continuity of $f$ implies the $f_n$ are equicontinuous and their lifts to $\mathbb{R}$ are uniformly bounded, and then it follows that their linear nullhomotopies are also equicontinuous). Thus $p$ is injective and hence an isomorphism.