Variant Of Inscribed Square Problem Looking at the inscribed square problem, I noticed that for some continuous curves $c:S^1\to\mathbf{R}^2$, one can prove that every continuous curve $c'$ with $\|c-c'\|_\infty<\epsilon$ contains an inscribed square.
For example, take $c$ to be the equilateral triangle (oriented like the letter $\Delta$) with bottom edge equal to $[0,1]\times\{0\}$.  Then choose $0<a<1/2$ such that $1-a<a\sqrt{3}$ (meaning that a square sitting on $((a,0),(1,0))$ has its left upper vertex still inside the triangle) and define $f,g:[0,1]^2\to\mathbf{R}$ by letting $f(t,s)$ denote the signed distance of the upper left vertex of the square sitting on $((at,0),(1-as,0)$ from the curve $c$ (if it sits inside of $c$, make this term negative) and similar let $g(t,s)$ denote the signed distance of the upper right vertex from $c$.  Now $f(0,s),g(t,0)>0$ but $f(1,s),g(t,1)<0$ for all $s,t\in[0,1]$ so there should exist a point $p\in[0,1]^2$ with $f(p)=g(p)=0$. (What's exactly the formal argument here?)  And for $\epsilon$ sufficiently small, this works with $c'$ as well (where we must work with an appropriate parameterization of the bottom edge).
Which other results exist of this kind?  Also, would it work if we don't assume that $\|c-c'\|_\infty<\epsilon$ but only that $c'$ lives in the $\epsilon$-tube around $c$ and there is a homotopy between $c$ and $c'$ inside this tube?  And what features can I use to infer that two continuous functions $f,g:[0,1]^2\to\mathbf{R}$ have a common zero?
[EDIT] Yep, think one can show it for the homotopy case as well.  Also, the title of that question is misleading, it's not a variant of the problem, more like a solution on a certain class of curves...
 A: 
what features can I use to infer that two continuous functions $f,g:[0,1]^2\to\mathbf{R}$ have a common zero?

A general result is the main theorem of Part II at p. 86 of [CS].

Let $f:D\to P$ be a mapping of a disk into the plane, let $C$ be the boundary circle of $D$, and let $y$ be a point of the plane not on $fC$. If the winding number of $f|C$ about $y$ is not zero, then $y\in fD$; i.e. there is a point $x\in D$ such that $fx=y$.

But whereas the calculation of the winding number is a technical task, your special case follows from a known theorem in dimension theory. 
Let $A$ and $B$ be disjoint subsets of a topological space $X$. A closed set $L\subset X$ is a *partition between $A$ and $B$ if there exist open sets $U,V\subset X$ such that $A\subset U$, $B\subset V$,  $U\cap V=0$, and $X\setminus L = U\cup V$.
1.8.1.  Theorem.  Let  $A_i$ and  $B_i$, where  $i = 1,  2, \dots,  n$, be  the  subsets  of the $n$-cube $I^n$  defined  by  the  conditions $A_i = \{\{x_j\}\in I^n:  x_i = 0\}$  and $B_i = \{\{x_j\}\in I^n:  x_i = 1\}$, i.e.,  the pairs  of  opposite faces of $I^n$. If  $L_i$ is a  partition  between  $A_i$  and $B_i$  for  $i = 1, 2,\dots, n$ then $\bigcap_{i=1}^n L_i\ne \emptyset.$ 
Proof.  Let us consider open sets $U_i, W_i\subset  I^n$  such  that $A_i\subset  U_i$,  $B_i\subset W_i$,  $U_i\cap W_i =
\emptyset $ and  $I^n\setminus L_i  = U_i\cup W_i$  for  $i = 1,2,\dots, n$.  Since  $(I^n\setminus W_i)\cap (I^n\setminus U_i)=
I^n\setminus (U_i\cup W_i)=L_i$, the formulas 
(1) $f_i(x)= \begin{cases}
\frac 12\frac {\rho(x,L_i)}{ \rho(x,L_i)+ \rho(x,A_i)}+\frac 12\mbox{ for }x\in I^n\setminus W_i\\
-\frac 12\frac {\rho(x,L_i)}{ \rho(x,L_i)+ \rho(x,B_i)}+\frac 12\mbox{ for }x\in I^n\setminus U_i 
 \end{cases},$
define  for $i = 1,2,\dots,  n$  a  continuous  function  $f_i: I^n\to I$ [Here $\rho$ is the distance. AR]. Clearly,  we have 
(2)  $f_i^{-1}(1/2) = L_i$, $f_i(A_i)  =\{1\}$  and  $f_i(B_i) =\{0\}$. 
Assume  that  $\bigcap_{i=1}^n L_i =\emptyset$;  it follows from the first part of  (2)  that  the continuous mapping  $f: I^n\to I^n$  defined by  letting  $f(x)  = (f_1(x), f_2(x),\dots,  f_n(x))$  for $x\in I^n$ does  not  assume the  value  $a = (1/2, 1/2,\dots , 1/2) \in  I^n$. The composition $g: I^n\to I^n$ of the mapping $f$ and  the  projection $p$  of  $I^n\setminus\{a\}$  from  the  point  $a$  onto  the  boundary  of  $I^n$,  i.e.,  onto the  set  $B  =\bigcup_{i=1}^n (A_i\cup B_i)$, satisfies  the  inclusion  $g(I^n)\subset B$;  by  the  second and  the third  part  of  (2),  we  have $g(A_i)\subset B_i$,  and $g(B_i)\subset A_i$. The  last three  inclusions show  that  $g(x)\ne x$  for  every  $x\in I^n$, which  contradicts the Brouwer fixed-point  theorem.  Hence  $\bigcap  L_i\ne\emptyset$. $\square$

Now $f(0,s),g(t,0)>0$ but $f(1,s),g(t,1)<0$ for all $s,t\in[0,1]$ so there should exist a point $p\in[0,1]^2$ with $f(p)=g(p)=0$.

Now the required claim follows from the theorem for $n=2$, $L_1=g^{-1}(0)$, $U_1=g^{-1}(0,\infty)$, $W_1=g^{-1}(-\infty,0)$, $L_2=f^{-1}(0)$, $U_2=f^{-1}(0,\infty)$, and $W_2\in f^{-1}(-\infty,0)$.
References 
[CS] W.G. Chinn, N.E. Steenrod. “First concepts of topology”, Random House, New York, 1966?.
[Eng] Ryszard Engelking, Dimension  Theory,  North-Holland, Amsterdam and  Polish  Scientific Publishers, Warsaw, 1978.
