This looks like a classic application of positive lookbehind, but unfortunately perl doesn't support that. In fact, doing this (matching the preceding text of a character in a string with a full regex whose length is indeterminable) can only be done with .NET regex classes, I think.
However, positive lookahead supports full regexes, so all you need to do is reverse the string, apply positive lookahead (like unicornaddict said):
perl -pe 's/(.)(?=.*?\1)//g'
And reverse it back, because without the reverse that'll only keep the duplicate character at the last place in a line.
MASSIVE EDIT
I've been spending the last half an hour on this, and this looks like this works, without the reversing.
perl -pe 's/\G$1//g while (/(.).*(?=\1)/g)' FILE_NAME
I don't know whether to be proud or horrified. I'm basically doing the positive looakahead, then substituting on the string with \G specified - which makes the regex engine start its matching from the last place matched (internally represented by the pos() variable).
With test input like this:
aabbbcbbccbabb
EFAUUUUH
ABCBBBBD
DEEEFEGGH
AABBCC
The output is like this:
abc
EFAUH
ABCD
DEFGH
ABC
I think it's working...
Explanation - Okay, in case my explanation last time wasn't clear enough - the lookahead will go and stop at the last match of a duplicate variable [in the code you can do a print pos(); inside the loop to check] and the s/\G//g will remove it [you don't need the /g really]. So within the loop, the substitution will continue removing until all such duplicates are zapped. Of course, this might be a little too processor intensive for your tastes... but so are most of the regex-based solutions you'll see. The reversing/lookahead method will probably be more efficient than this, though.