Is there a set $S\subseteq [0,1]$ with $|S|=2^{\aleph_0}$ and distinct pairwise distances?

Consider $\Bbb R$ as a vector space over $\Bbb Q$, choose (Zorn) a basis $S$ consisting (after rescaling with rational scalars) of elements in $(0,1)$. Then we have no relation of the shape $\pm(s_1-s_2)=(s_3-s_4)$ for $s_1,s_2,s_3,s_4\in S$. The cardinality of the basis is also the required one.

You can actually make $S$ a (copy of the) Cantor set.

First a claim: if $[a_1,b_1]$, $[a_2,b_2]$, ..., $[a_n,b_n]$ is a finite set of intervals, ordered left-to-right ($b_1<a_2$, $b_2<a_3$, etc) then we can choose points $p_{2i-1}<q_{2i-1}<p_{2i}<q_{2i}$ in $(a_i,b_i)$ with the following property: if $i$, $j$, $k$ and $l$ are distinct and $x\in[p_i,q_i]$, $y\in[p_j,q_j]$, $z\in[p_k,q_k]$, and $w\in[p_l,q_l]$ then six possible distances are distinct. To prove this draw inspiration from the first answer and choose $x_{2i-1}<x_{2i}$ in $(a_i,b_i)$, such that $X=\{x_j:j\le 2n\}$ is linearly independent over $\mathbb{Q}$. Then $d$ is injective on $[X]^2$ and by continuity we can choose the intervals $[p_j,q_j]$ to preserve this for points taken from distinct intervals.

Given this claim follow the common construction of the Cantor set where at stage $n$, instead of deleting the middle thirds of all $2^n$ intervals, you take the next $2^{n+1}$ intervals using the claim.

The resulting Cantor set is as required: given two distinct unordered pairs $\{x,y\}$ and $\{z,w\}$ find a stage where distinct points are in distinct intervals; it follows that $|x-y|\neq|z-w|$.

The existence of such a Cantor set also follows from Mycielski's theorem on Independent sets in topological algebras.