We only need to consider word pairs - i.e. pairs of words such that each is the other reversed. Additionally, we only need to consider word pairs where the three letters used are each unique - so eel/lee, for example, would be eliminated. From these two properties, there are only 216 word pairs that need to be considered. Sandor's list of letters in the middle of words, it would seem, does not take into account the second property. When that property is taken into account, the letters in the middle of words are as follows: ABDEHIORTUVWY. From there, only 47 word pairs could fit in the middle. From these 47 words, only 6 letters are a part of multiple word pairs and so could be in the center of the square: AEIORU. Let us consider each of these.
Case I: The word pairs in this case are air/ria, bid/dib, dit/tid, and div/vid. Of the four, only air/ria does not contain the letter D, and so this must be one of the two word pairs, and the other must contain a D:
However, the only word pair with a D in the center is ado/oda, which shares A with air/ria. Therefore, I cannot be the center letter.
Case R: The word pairs in this case are arb/bra, are/era, bro/orb, and bru/urb. Much like in case I, one word pair is are/era and the other must contain a B:
However, the only word pair with a B in the center is abo/oba, which shares A with are/era. Therefore, R cannot be the center letter.
Case U: The word pairs in this case are bud/dob, bur/rub, but/tub, and duo/oud. As in the previous cases, one word pair is duo/oud and the other must contain a B:
However, the only option for the top row, abo/oba, shares o with duo/oud. Therefore, U cannot be the center letter.
Case E: There are six word pairs in this case: bed/deb, dew/wed, her/reh, hey/yeh, tew/wet, and wey/yew. The pair bed/deb can immediately be eliminated, since b can only intersect abo/oba and d can only intersect ado/oda. Similarly, dew/wed can only intersect ado/oda and owt/two, and so can be eliminated immediately. Therefore, one word pair must be her/reh or hey/yeh, and the other must be tew/wet or wey/yew. W must intersect owt/two, and the only word pair that intersects H that has an O is mho/ohm:
The WE? must now be wey/yew, but the only word pair with y in the middle is nys/syn, which does not contain M. Therefore, E cannot be the center letter.
Case O: There are ten word pairs in this case. The letters BDW each can only intersect a word pair with an O in it (abo/oba, ado/oda, and owt/two, respectively), which eliminates every word pair except for rot/tor. Therefore, with only one word pair not immediately eliminated, O cannot be the center letter.
Case A: There are fourteen word pairs in this case. The letters BDTV can be eliminated since the word they intersect with must contain an A (abo/oba, ado/oda, ate/eta, and avo/ova, respectively). Eliminating those leaves four word pairs: hay/yah, raw/war, ray/yar, and way/yaw. The two word pairs used must be one with a Y, and raw/war:
However, W can only intersect owt/two, and Y can only intersect nys/syn. Those two word pairs share no letters, and so cannot simultaneously be put in place. Therefore, A cannot be the center letter.
As all six possibilities have been eliminated, no letter can be in the center, and so the square is impossible.