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
 A: The relevance of the fact that the set is closed is that the limit point must be in the set. For example $\sin (1/(z+1))$ has infinitely many zeros in the open unit disc, but is not zero. (The sole point of accumulation is $-1$, outside the domain, and the function is not analytic there.)
On the multiplicity: first I think distinct zeros are meant. Second, a non-zero analytic function (on a connected domain) cannot vanish to infinite order anywhere; this would give the series there is $0$ and so the function is zero.   
A: Since the disk $\mathcal{D}$ is bounded and closed it is compact, so, by Heine-Borel it fulfils the Bolzano-Weierstrass property: every infinite set has at least one limit point. So the set of $\mathcal{Z}$ of zeros has a limit point, and you're right to go with the identity theorem. So you can see the reason for closure: it lets you call on the Heine-Borel theorem.
Incidentally, before about 1930 and Pavel Alexandrof's school, "compact" meant by definition having the Bolzano-Weierstrass property.
A: 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). 
