Here is a construction of a polygon that cannot fall through the hole.

Begin with a regular $MN$-gon circumscribed around a unit circle, where $M\gg N\gg 1$ and $N$ is even. For every $M$th side, draw a segment of length 1 extending this side in the clockwise direction. Take a rectangular neighborhood of width $\varepsilon\ll 1/MN$ of each of these segments. The manhole is the union of the $MN$-gon and these $N$ narrow rectangles.

Suppose that the polygon can be moved through the hole. Then there is a moment when its center is on the ground level. Consider the intersection line of the two planes: the ground and the polygon. Its intersection $S$ with the polygon consists of the central segment of length $>2$ and several pairs of short segments arising from the pairs of symmetric narrow rectangles ($N$ should be chosen large enough so that there are at least 3 such pairs for any line through the center).

The same line on the ground must intersect the hole by a set that contains $S$ in the interior. The central segment must go very close to the center of the hole because there is no other place for a segment of length 2 (we can control how close is should be by choosing $M$ large enough). Now consider a symmetric pair of short segments in $S$ somewhere in the middle (to avoid borderline effects). If the line on the ground is sufficiently close to the center (and we can ensure that), this pair of segments must fit into a pair of symmetric rectangles in the hole. But it is easy to see that it cannot fit - locally these rectangles are just parallel strips of width $\varepsilon$ separated by a fixed distance $2-\varepsilon$, and however you rotate the intersecting line, you will get either shorter segments or shorter interval between them.