A subset of a rectangle that 'blocks' every curve that goes from right to left must connect upper and lower sides

I believe your assertion is false. Let $A=\gamma_1([0, \infty)) \cup \gamma_2([0, \infty))$ be a union of two curves that spiral in towards the disk $D_r$ of radius $r$ but never reach it. Something like this:

Each closure $\overline{\gamma_i([0, \infty))} = \gamma_i([0, \infty)) \cup \partial D_r$ is disjoint from the other curve, so the two curves are a separation of $A$; neither component of $A$ meets the top and bottom edges.

Now suppose (for contradiction) that some curve $f \colon [0, 1] \to [-1,1]^2 \setminus A$ has $f(0)$ on the left edge and $f(1)$ on the right edge. By compactness $f$ attains some minimum distance $q$ away from the center. We can't have $q>r$ because $[-1, 1]^2 \setminus A \setminus D_q$ then certainly does not have such a path $f$. We can therefore let $t_0 = \inf\{t \in [0, 1] : f(t) \in D_r\}$, so then $f(t_0) \in \partial D_r$ and $f([0, t_0))$ is disjoint from $D_r$. Choose a small $\epsilon < r$. By continuity, there is some $\delta > 0$ with $f([t_0-\delta, t_0))$ all $\epsilon$-close to $f(t_0)$. However, since $f(t_0 - \delta) \notin D_r$, we can see that $f$ must always do another loop around $D_r$ on $(t_0-\delta, t_0)$ before being able to reach $f(t_0)$. But another loop around is impossible because $\epsilon$ is small, so we have a contradiction, and we can conclude that any path from the left edge to the right edge must meet $A$.