If $h:[0,1]^2\to R$ is a continuous function that is positive on $\{x_1=0\}$ and negative on $\{x_1=1\}$, does $h^{-1}(0)$ contain a continuous curve Let $h:[0,1]^2\to\mathbb{R}$ be a continuous function such that $h<0$ on $\{0\}\times[
2/5,3/5]\cup [2/5,3/5]\times\{0\}$ and $h>0$ on $\{1\}\times[2/5,3/5]\cup [2/5,3/5]\times\{1\}$. By the Intermediate Value Theorem, there exists a function $\xi:[2/5,3/5]\to [0,1]$ such that $h(\xi(t),t)=0$ for each $t\in[2/5,3/5]$. Is possible to choose $\xi$ in a way that $\xi$ is continuous?

Not always.
Consider the curve 
Define $h(x,y)$ to be the lowest distance from $(x,y)$ to any point on the curve, if $(x,y)$ lies to the upper-right of the curve, and the opposite of that distance if $(x,y)$ lies to the lower-left of the curve. $h$ then meets all of your conditions, but $\xi$ cannot be chosen to be continuous at $\frac 12$.
