Can a non-constant analytic function have infinitely many zeros on a closed disk? I think not, however my proof is quite sketchy so far..
My attempt: Suppose an analytic function f has infinitely many zeros on some closed disk D. Then there exists a sequence of zeros in D with a limit point in D. Thus by the identity theorem (Let D be a domain and f analytic in D. If the set of zeros Z(f) has a limit point in D, then f ≡ 0 in D.), f is identically zero and thus constant.
My main reasons for confusion (other than having a weak understanding of the identity theorem):
-Couldn't such a function f have a finite number of distinct zeros, each with infinite multiplicity? in this case there wouldn't be a convergent sequence of zeros...
-What is the relevance of the fact that D is closed?
Any help in understanding this problem would be greatly appreciated! 
Thanks

Your proof is correct, you just need to realize that when you say "has infinitely many zeros" you mean "has infinitely many points where it evaluates to $0$", so one is not talking about multiplicities here. The importance of $D$ being a closed disk is that it is then compact, and that implies the existence of a convergent sequences of zeros of the functions, which allows you to invoke the identity theorem (which you certainly do want to understand fully). 
