Well, this is what I've come up with. When white pushes out, black needs to get a little thickness first (otherwise, white can squiggle her way to freedom).
Alex Weldon: It's probably better to push once more. W still won't have enough eyespace to live.
If white cuts here at , then black must atari at (an atari at will cause big headaches for black). When white extends to , black must extend to . I believe black is well equipped to handle white's next step (three of which I've explored, none of which seem to work).
Warfreak2: "then black must atari at (an atari at will cause big headaches for black)". Do not automatically atari every cutting stone! Black need not atari at all, just extend, to get liberties.
If white tries to go after the two black stones on the left, black has a series of ataris that will keep him safe. If white tries to go after the ones at the top, and take care of things. (Neither can white extend up at the top; because of the marked black stone, this atari-capture move is useful for black more than once.) Note that white cannot atari at a for the same reason.
At this point (or at least after he plays at a), black has enough liberties to overcome white's four in the corner.
If white moves up with , then black must be careful and extend first with . Otherwise, white can capture those two and be safe (the atari-capture moves don't work if black omits this move). After , black -- again -- has enough liberties to win the semeai.
I will now step aside and let others show me all my mistakes.
-- Scartol
Arno: one mistake I found (unless I am mistaken :-)
Note: should be at 'a' to reduce White's liberties, but anyway. After the sequence to White plays 'b' and is safe. Black cannot cut at 'c' and therefor White is safe. Also, 'd' is sente, as there is no ko threat to be found anywhere.
So is the stone alive?
Mef: Unless I'm misreading, at the very least after 'a' and 'b' are miai. If then white at 'c' should escape and live easily.