I've been thinking of this problem but I just can't complete a proof, here's a possible route (hopefully) towards a solution: By a simple continuity argument there is an such that on we have that does tend to zero. Now assume the following is true
Claim: If on the line with then on
then by an argument identical to the first on so the set on which has to be . I'm having a little trouble wih the claim though (particularly estimating in the verticla boundary of the set for some ; this is enough by the MMP).
Sorry I can't be of more help, if you find a proof please post it here.
There's probably a nicer way than this (i.e. a way that invokes maximum modulus at least). But for now I don't see why this doesn't work.
Take the sequence of points . This sequence is in D.
By continuity of f we can flex the limit in and out of the function:
Now f is holomorphic and bounded on D, so it has no singularity at infinity (in particular, no erratic essential singularity behavior). So
exists. So all unbounded sequences in D tend to this limit. We just showed one such sequence tends to A.
(So A is what all unbounded sequences in D tend to. In particular, any unbounded sequence with fixed imaginary component between 0 and 1 tends to A.)