= Solution
Write $\beta(s)=(\beta_1(s),\beta_2(s))$, $\gamma(t)=(\gamma_1(t),\gamma_2(t))$, and
$$
h(s,t)=\beta(s)-\gamma(t).
$$
The endpoint conditions imply
$$
\begin{array}{ll}
h_1(s,-1)\geq0,&h_1(s,1)\leq0,\\
h_2(-1,t)\leq0,&h_2(1,t)\geq0.
\end{array}
$$
Let $P(x)=\max(-1,\min(1,x))$ and define the continuous self-map of $I^2$
$$
F(s,t)=\bigl(P(s-h_2(s,t)),\ P(t+h_1(s,t))\bigr).
$$
By Brouwer, $F$ has a fixed point $(s,t)$. If $s$ is interior, its fixed-point equation gives $h_2=0$. At $s=-1$ or $s=1$, the projection equation and the corresponding displayed boundary sign again force $h_2=0$. The identical argument in the second coordinate gives $h_1=0$. Hence
$$
h(s,t)=0,
$$
so
$$
\boxed{\beta(s)=\gamma(t).}
$$
The two paths intersect.
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