There exists a probability space $(\Omega,\mathcal{A},P)$ sucht that:

 1. For each $x\in[0,1]$ there is a random variable $B_x:\Omega\rightarrow\{0,1\}$ with $P(B_x=1)=p$.

 2. The collection $\{B_x,x\in[0,1]\}$ is independant.

The construction is standard: For $x\in[0,1]$ let $$(\Omega_x,\mathcal{A}_x,P_x):=(\{0,1\},\mathcal{P}(\{0,1\}),\text{Bernoulli}(p))$$
be the probability space modelling the toss of a $p$-coin. Then you define
$$\Omega:=\prod_{x\in[0,1]}\Omega_x,~~\mathcal{A}:=\bigotimes_{x\in[0,1]}\mathcal{A}_x,$$
where $\bigotimes$ means taking the product sigma field. You can then show that the product measure $P=\bigotimes_{x\in[0,1]}P_x$ exists. The random variables $B_x$ are nothing but the projections. Your random set $S$ would be defined as
$$S:=\{x\in[0,1]~:~B_x=1\}.$$
This random set behaves 'badly' (in some sense its not measurable with probability $1$). See
https://terrytao.wordpress.com/2007/06/18/the-lebesgue-differentiation-theorem-and-the-szemeredi-regularity-lemma/, 5th paragraph.

Subsets $S\subseteq[0,1]$ that behave like the random set you have in mind should have properties like $\lambda([0,t]\cap S)=pt$ for each $t\in[0,1]$, where $\lambda$ is lebesgue measure. You can show that this can not be done with a measurable set $S$.