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?

connectedsubset of $[0,1]^2$ with a full projection $\endgroup$9more comments