Two loops are intersecting each other in convex set of R In the figure bellow  I was trying to prove that $ γ_1$ is intersecting $ γ_2$ at some point inside the convex and compact subset of $R$, then I thought to join $x, y$ and $r, z$ by line segments $s_1$ and $s_2$. to make a loop considering by $ γ_1$ $, γ_2$ and the line segments $s_1$ and $s_2$ and maybe some parts of the boundary of the space.


The idea I was working on is to assume that the two paths are not intersecting and hence the shape of the loop we constructed above is homotopic to the circle $S^1$ so if one takes any loop on the main loop he should see that it has a generator and then this contradicts the fact that the fundamental group of $S^1$ is $Z$ which has no generators.. Can I do that? or are they some other methods using homotopy or/and fundamental group to prove this situation?



 A: EDIT: After reconsidering my original idea in order to elaborate on the crucial point I came to the conclusion that in the end one would run into awful details there, much alike to re-inventing the Jordan curve theorem. It's possible to stow away the awful details right from the beginning by invoking Jordan's curve theorem directly. In the context of correct application of Jordan, one should avoid using intuition, so let's be very explicit:
Given $\gamma_2\colon[0,1]\to [a,b]\times ]a,b[$ with $\gamma_2(0)=(a,y_1)$ and $\gamma_2(1)=(b,y_2)$ define the  closed curve $\gamma\colon[0,4]\to\mathbb R^2$ by
$$\gamma(t)=\begin{cases}
\gamma_2(t)&\text{if }0\le t\le 1,\\
(b,y_1)+(t-1)\cdot(1,y_1-a+1)&\text{if }1\le t\le 2,\\
(b+1,a-1)+(t-2)(b-a+2,0)&\text{if }2\le t\le 3,\\
(a-1,a-1)+(t-3)(1,y_1-a+1)&\text{if }3\le t\le 4.\end{cases} $$
Since $\gamma_2$ is simple, $\gamma$ is a simple closed curve and hence it complement in $\mathbb R^2$ has two connected components, a bounded interior  and an unbounded exterior component and the curve is the boundary of both.
The rectangles $$A=[a,b]\times[b,+\infty[,\qquad B=[a,b]\times]-\infty,a-1[,\qquad C=[a,b]\times]a-1,a]$$ are all disjoint to $\gamma$.
As they are connected, they are each subset of either the bounded or the unbounded component. As $A,B$ themselves are unbounded, they belong to the unbounded component.
All point sufficiently close to the point $(\frac{a+b}2,a-1)$, say at distance$<\min\{1,\frac{b-a}2\}$, are either on $\gamma$, or in $B$, or in $C$. As point on the curve, $(\frac{a+b}2,a-1)$ is on the boundary of the interior, hence $C$ must be in the interior.
Therefore, any curve $\gamma_1$ with endpoints $(x_1,a)\in C$ and $(x_2,b)\in A$ must intersect $\gamma$ at some point $\gamma(t)$ with $0\le t<4$. If additionally $\gamma_1$ resides in $[a,b]\times[a,b]$, we conclude $t\le 1$, i.e. $\gamma_1$ and $\gamma_2$ intersect.
