Bedevil your hangman opponent
Python 2, 60 bytes

lambda l:min(l,key=lambda x:sorted(map(`l`.count,set(x)))+l)

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Explanation
The idea is much like every other answer. In every word, we compute the frequency of each character. Then we take the word whose least frequent character occurs most often.
In the code, we sort the array as opposed to finding the minimum. Here is an example to show how this works: Input: thee lee baa. For each word, we compute a frequency table (duplicates are ignored): [(1, 1, 4), (1, 4), (1, 2)]. After sorting, the lexiocgraphically smallest list is (1, 1, 4), therefore the answer is thee.
There is also an edge case to consider: Input: a bc, which gives the following frequency table: [(1), (1, 1)]. The lexicographically smallest list would be (1), which is incorrect since the second list contains more 1s. I'll leave out the specifics of how the code resolves this issue