If $S⊂[0,1]^2$ intersects every connected subset of $[0,1]^2$ with a full projection on the $x$-axis, must $S$ have a connected component with a full projection on the $y$-axis?
An equivalent form：
If $S⊂[0,1]^2$ intersects every connected subset of $[0,1]^2$ with a full projection on the $x$-axis and $T⊂[0,1]^2$ intersects every connected subset of $[0,1]^2$ with a full projection on the $y$-axis, must $S\cap T\neq \emptyset$?
The motivation of this question:
The question came to me when I thought about the Brouwer fixed-point theorem:
Let $f=(f_1,f_2)$ be a continuous function mappping $[0,1]^2$ to itself. Then
$$S\triangleq\{(x,y)\in[0,1]^2:f_1(x,y)=x\}$$
intersects every connected subset of $[0,1]^2$ with a full projection on the $x$-axis and
$$T\triangleq\{(x,y)\in[0,1]^2:f_2(x,y)=y\}$$
intersects every connected subset of $[0,1]^2$ with a full projection on the $y$-axis.
My further question:
If we assume that $S\subset [0,1]^2$ is a close set, what is the answer to my question, that is, if a close set $S⊂[0,1]^2$ intersects every connected subset of $[0,1]^2$ with a full projection on the $x$-axis, must $S$ have a connected component with a full projection on the $y$-axis?
Best Answer
A counterexample to this statement was posted as a comment by Dejan Govc to the Math StackExchange question, Do partitions of a square into two sets always connect one pair of opposite edges?.
For $0 < r < \tfrac{1}{2}$, let $S_r$ be the boundary of the square $\bigl[\tfrac{1}{2}-r,\tfrac{1}{2}+r\bigr]\times \bigl[\tfrac{1}{2}-r,\tfrac{1}{2}+r\bigr]$, and let $$ S = \{(0,0),(1,0),(0,1),(1,1)\} \;\;\cup \bigcup_{r\in \mathbb{Q}\cap (0,1/2)} S_r. $$ Note that no connected component of $[0,1]^2\setminus S$ has full projection onto the $x$-axis, and therefore any connected subset of $[0,1]^2$ with full projection onto the $x$-axis must intersect $S$. However, no connected component of $S$ has full projection onto the $y$-axis.