Self-complete set in square Consider $Q = [0,1]^2$ and $\phi:Q\to \mathbb{R}_{\geq 0}$ such that $\phi$ is a Lipschitz continuous function with a constant, say $\lambda_\phi$. 
For each $x\in[0,1]$ we put $S(x) = supp \phi(x,\cdot)$ - i.e. $S(x)$ is a closure of the following set
$$
(y\in[0,1]:\phi(x,y)>0).
$$
We call a non-empty set $A\subseteq [0,1]$ self-complete if for any $x\in A$ holds $S(x)\subseteq A$. Could you give some ideas how to verify if there are self-complete sets? Any ideas are more than welcome.
What I was able to prove is that $A$ is a set of a positive measure and if there is a self-complete set $A$ then its closure is also self-complete.
P.S. The term self-complete I use myself - maybe there are intersection in a terminology.
Edited: the problem can be reformulated as following: given an open set $G\subset Q$ verify if there is a closed non-empty set $A\subset [0,1]$ such that if $x\in A$ and $(x,y)\in \bar{G}$ then $y\in A$. 
 A: 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. 
