Proving that these curves intersect Let $\Gamma$, $\Sigma$ be two curves with ranges in $(\{0\}\cup\mathbb{R}_{+})^2$. $\Gamma$ starts on the $y$ and ends on the $x$ axis: $\Gamma(0)=(0,\gamma_2),\Gamma(1)=(\gamma_1,0)$. $\Sigma$ is a "positive unit step" curve, in the sense that it is a concatenation of horizontal and vertical segments of unit length in the positive $x$ or $y$ directions (it "moves" by a unit step either to the right or upwards). Moreover, $\Sigma$ starts on one of the axes, and, if $\Sigma(0)=(\sigma_1,\sigma_2)$, then $\sigma_j\le \gamma_j$.
The following statement is clear, but I haven't proved it rigorously yet: $\Gamma$ and $\Sigma$ intersect. Consider the closed curve formed by $\Gamma$, the origin, and the two segments on the axes in between. If a point moving along $\Sigma$ is in the (bounded) region enclosed by this closed curve, it will eventually exit it, since $\Sigma$ is unbounded.
Could someone please assist me in constructing a rigorous proof?
 A: You are pretty close. As you said, consider the closed curve $\Lambda$ which starts at the origin, goes straight to $(0, \gamma_2)$, follows $\Gamma$, and then goes straight from $(\gamma_1, 0)$ to the origin. By the Jordan curve theorem, this divides $\mathbb{R}_{\geq 0}^2$ into two regions. Let us call the bounded region $A$ and the unbounded region $B$, and note $\mathbb{R}_{\geq 0}^2 = A \cup \operatorname{im}(\Lambda) \cup B$, which are disjoint. Also note the Jordan curve theorem says that $\partial A = \operatorname{im}(\Lambda)$.
We now construct the regions $X$ and $Y$, where $X = A \cup \operatorname{im}(\Lambda) - \operatorname{im}(\Gamma)$, and $Y = B$. It is easy to see that we have a partition $\mathbb{R}_{\geq 0}^2 = X \cup \operatorname{im}(\Gamma) \cup Y$ and that $\partial X = \operatorname{im}(\Gamma)$. Now we know that $\Sigma$ starts in $X \cup \operatorname{im}(\Gamma)$ and ends up in $Y$ from the givens.
If $\Sigma$ starts within $\operatorname{im}(\Gamma)$ we are done. So let us assume $\Sigma(0) \in X$. Since $\Sigma$ leaves $X$, we know $\exists s \in \mathbb{R}$ s.t. $\Sigma(s) \in \partial X$. But $\partial X = \operatorname{im}(\Gamma)$, so $\Sigma(s) \in \operatorname{im}(\Gamma)$, providing us with our intersection.
