**Proposition.** Suppose that a subset $X\subset R^3$ is  semilocally simply connected (SLSC). Then 
for each $x\in X$, the group $G= \pi_1(X,x)$ is torsion-free. 

Proof. I will be using the fact that if $U$ is an open connected subset of $R^3$, then $\pi_1(U)$ is torsion-free, see for instance [here](https://math.stackexchange.com/questions/680998/is-there-a-domain-in-mathbbr3-with-finite-non-trivial-pi-1-but-h-1-0/684883#684883). 

Let $c: S^1\to X$ be a loop representing an element of order $n$ in $\pi_1(X,x)$. Accordingly, let $c_n: S^1\to X$ be the precomposition of $c$ with the map $z\mapsto z^n$ and $h: D^2\to X$ be 
an extension of $c_n$ to the unit disk. The image $Y:= h(D^2)$ is compact and locally path-connected. Thus, there is a function $\phi(\delta)$ such that if $d(y_1, y_2)\le \delta$, $y_1, y_2\in Y$, then there is a path of diameter $\le \phi(\delta)$ in $Y$ connecting $y_1, y_2$. Furthermore, since $Y$ is compact and $X$ is assumed to be SLSC, there exists  $\epsilon>0$ such that if $\alpha: S^1\to Y$ has image of diameter $\le \epsilon$, then 
$\alpha$ extends to a continuous map $D^2\to X$. 

Now, consider the system of open $\frac{1}{i}$-neighborhoods $U_n$ of $Y$ in $R^3$. Let $r: U_i\to Y$ denote a (likely discontinuous) nearest-point projection (defined via the Axiom of Choice). 
Since each group $\pi_1(U_i,x)$ is torsion-free, the map $c: S^1\to Y$ extends to a (continuous) map $f_i: D^2\to U_i$. 

I will now imitate the standard argument (I think, due to Borsuk), from the proof that each finite-dimensional compact locally-contractible metrizable space is ANR. 

Given a triangulation $T$ of $D^2$, I define the map $g_i: T^{(0)}\to Y$ as the composition of the restriction of $f_i$ to the $T^{(0)}$ with the projection $r$. The goal is to show that for large $i$, the map $g_i$  extends to a map $D^2\to X$ which restricts to $c$ on $S^1=\partial D^2$. 

First of all, if $\frac{1}{i}\le \delta$, and $T$ is such that the diameters of the images under $f_i$ of 
the edges of $T$ are $\le\delta$, then for each edge $e=[v,w]$ of $T$, $d(g_i(v), g_i(w))\le 3\delta$. 
Hence, by the local path connectivity of $Y$, we can extend $g_i$ to $e$ so that $g_i(e)\subset Y$ has diameter $\le \phi(3\delta)$. In the case of boundary edges of the disk $D^2$, we will assume that $g_i|_e=c|_e$.  Note that for each 2-simplex $\Delta$ in $T$, the diameter of $g_i(\partial \Delta)$ is $\le 3\phi(3\delta)$. By taking $i$ sufficiently large and taking the triangulation 
$T$ sufficiently fine, we can assume that $3\phi(3\delta)\le \epsilon$, where $\epsilon$ is defined as above. Hence, for this value of $i$, the map $g_i$ extends to a map $g: D^2\to X$. It follows that $[c]=1\in G=\pi_1(X,x)$ and, hence, $G$ is torsion-free. qed