Several different thoughts. I take it that a set $A \subseteq [0,1]$ is self-complete if $\bigcup_{x \in A} S(x) \subseteq A$, where $S \colon [0,1] \to P([0,1])$ is some function defined as above in terms of a single fixed Lipschitz $\phi\colon [0,1]^2 \to \mathbb{R}_{\geq 0}$.
(1) In some cases the set will be $[0,1]$ itself, for example if $\phi(x,y) = 1$ then $[0,1]$ is the only (nonempty) self-complete set.
(2) Regardless what $S$ is, we can make a self-complete set by just creating a set that is "closed under $S$" in an appropriate sense. You simply use transfinite induction to work towards the goal that whenever $x \in A$, $S(x) \subseteq A$. It goes like this:
Pick any point $x_0$ and let $A_0 = \{ x_0\}$. Now, by transfinite induction, for $\lambda > 0$ let
$A_\lambda = \left ( \bigcup_{\kappa < \lambda} A_\kappa \right ) \cup \left ( \bigcup_{\kappa < \lambda}\, \bigcup_{x \in A_\kappa} S(x) \right ) $
This has the property that $A_\kappa \subseteq A_\lambda$ whenever $\kappa < \lambda$. We can't keep adding new points forever, because there are only as many points as the cardinality of [0,1]. So eventually we will have $A_\kappa = A_{\kappa+1}$, and this will be a self-complete set.
(3) Normally, you could replace this transfinite induction with a "top-down" argument. In fact the intersection of any family of self-complete sets would be self complete except for the requirement you added that self-complete sets must be nonempty. Without that requirement, each $\phi$ would be associated with a particular minimal self-complete set. The argument in part (2) above shows that for any $x \in [0,1]$ there is a nonempty self-complete set containing $x$. The intersection of all such sets will still be a self-complete set containing $x$, and so will be a minimal self-complete set among the ones that contain $x$. In fact, this is the set that was constructed in part (2).
(4) You cannot prove that every self-complete set is of positive measure. If $\phi$ is identically 0 then $S(x) = \varnothing$ for every $x$, and so every (nonempty) set is self-complete.