Solution:
Consider the following version of the problem:
A rectangle $R$ is divided into a set $S$ of finitely many smaller rectangles such that no three rectangles in $S$ share a common corner. For each $r \in S$, draw two non-intersecting arcs inside $r$, connecting the pairs of adjacent corners of $r$ (there are two ways to do this, by connecting either the horizontally or vertically adjacent corners). Prove that there does not exist a path from the bottom-left corner of $R$ to the top-right corner of $R$ by walking only along these arcs.

<center>Figure 1: A possible diagram of all the arcs. </center>

In this problem, consider an undirected graph where nodes correspond to corners of rectangles in $S$ and edges correspond to the arcs, connecting the two nodes that the arc connects. The degrees of the nodes corresponding to the corners of $R$ are all exactly 1. Since no three rectangles in $S$ share a common corner, all intersection points have a pattern like $\vdash$, $\neg$, $\perp$, or $\top$, so the degree of all other nodes is exactly 2. Therefore, this graph can be decomposed into several paths and cycles. The only possible endpoints of paths are the degree 1 nodes, which are the corners of $R$. It follows that if there exists a path from the bottom-left corner to the top-right corner of $R$, then there also exists a path from the bottom-right corner to the top-left corner of $R$. However, this is impossible because these two paths (viewed as planar curves inside $R$) must intersect, which cannot occur. Therefore, this claim is proved.

Returning to the original problem, suppose some operations were performed while choosing each rectangle at most once. Draw two arcs inside every rectangle, either both horizontal if the ant used this rectangle to move horizontally or both vertical if the ant used this rectangle to move vertically (or pick one arbitrarily if this rectangle was not used). By the new version of the problem, there does not exist a path from the bottom-left corner to the top-right corner of $R$, and it follows that it is impossible for the ant to have reached the top-right corner of $R$, finishing the proof.

Suppose that no rectangle was chosen in at least two operations. In particular, a rectangle cannot be selected in two consecutive operations.
At any point in the process, consider whether the last move by the ant was horizontal or vertical, and whether the most recently chosen rectangle was to the left or the right of the ant's path. In the first move, either the ant moved horizontally and the rectangle was to the left, or the ant moved vertically and the rectangle was to the right. We claim that this invariant is preserved throughout the entire process (see Figure 2 for a sample path). Assuming this claim, the final move to the top right corner must select the top right rectangle. If it is vertical, this is to the left of the path, and if it is horizontal, it is to the right of the path, both of which are impossible, providing a contradiction.

<center>Figure 2: A possible path by the ant. The red arrows are all vertical, with the corresponding rectangle to the right. The blue arrows are all horizontal, with the corresponding rectangle to the left. </center>

It remains to show that the invariant is preserved. Since four rectangles cannot intersect at a corner, each intersection has a pattern like $\vdash$, $\neg$, $\perp$, or $\top$.
First, assume the ant moves up, hence the chosen rectangle $r$ is on the right. The possible configurations are depicted in the first two diagrams in Figure 3.

<center>Figure 3: Possible ant moves going up or right. </center>

In each case, the ant must choose rectangle $s$ next (to avoid repeating $r$ twice), and we see that both side choices preserve a horizontal move with $s$ left, or a vertical move with $s$ right. By rotating the picture by $180^{\circ}$, we cover the two possibilities for the ant moving downward.
If the ant moves right, then $r$ must occur on the left, and the possible configurations are the last two diagrams of Figure 3. Once again, rectangle $s$ must be chosen next, and the invariant is similarly preserved. The case of the ant moving left is again handled by a $180^{\circ}$ rotation, completing the proof.

This is a variant of Solution 2. As in that proof, assume that no rectangle was chosen in two consecutive operations. We claim that for every move, the ant is in either the bottom-left or top-right corner of the chosen rectangle, and moves to the bottom-right or top-left corner.
This is clearly true of the first move. If the ant starts at the bottom-left or top-right corner on a move (choosing rectangle $r$), it is clear that they must move to the bottom-right or top-left of $r$. Assume they moved to the bottom-right corner of $r$, and chose rectangle $s$ in the next move. If they are at the bottom-right corner of $s$, then $r$ and $s$ are either equal or overlap, a contradiction. If they are at the top-left of $s$, then $r$ and $s$ intersect at a corner and no sides, and we must have 4 rectangles intersecting at a corner, again a contradiction.
Therefore they must be at the bottom-left or top-right corner of $s$, as desired. The case where the ant is at the top-left corner of $r$ is analogous.
If the ant is able to make it to the top right corner, their final move must select $r$ as the top-right rectangle, and they move to the top-right corner, which is therefore impossible.