If each $f_n$ has $b_n$ zeros, then the number of zeros of $g$ is at most $\liminf_{n \to \infty} b_n$ My question is the following:

Let $f_n:B_1(0) \to \mathbb{C}$ be holomorphic so that $f_n \to g$  uniformly on compact sets. Assume that $g$ is non-constant and let $b_n$ be the number of zeros (with multiplicity) that $f_n$ has. Show that the number of zeros of $g$ (with multiplicity) is at most $\underset{n \to \infty}{\liminf} b_n$.

My attempt:
Let $z \in B_1(0)$ be a zero of $g$.  Because $g$ is nonconstant, the zeros are isolated. So we can find an open ball about $z$ that contains no other zeros of $g$, say $B_\delta(z)$, for some $\delta > 0$.  Note $f_n \to g$ uniformly on $\overline{B_\delta(z)}$ so $\forall \epsilon > 0 \exists N$ such that $|f_n - g | < \epsilon$, $\forall n > N$.  For $\epsilon < \min_{w \in \partial B_\delta (z)} \{|g(w)| \}$ we see that $|f_n - g| < \min_{w \in \partial B_\delta (z)} \{|g(w)| \}$ for all $n > N$.  By Rouche's theorem, the number of zeros of $g$ on $B_\delta(z)$ is equal to the number of zeros of $g + (f_n - g) = g$ for all $n > N$.
I would like to then conclude that the number of zeros of $g$ on $B_1(0)$ is at most $\liminf_{n\to \infty} b_n$, but I am having trouble justifying this...
 A: Wlog we can assume $N=\underset{n \to \infty}{\liminf} b_n$ is finite (else nothing to prove as the zeroes of $g$ are at most countable as they are discrete in the open disc) and since the number of zeroes is always an integer, it follows there is an infinite subsequence of the $f_n$'s with precisely $N$ zeroes (counting multiplicities of course).
Passing to this subsequence we can assume wlog each $f_n$ has $N$ zeroes and we need to prove that $g$ has at most $N$ zeroes which follows now by contradiction as in the original post
(Let $M$ the number of zeroes of $g$, assume $M>0$ so there is something to prove and split them into distinct numbers $z_1,..z_k$ with multiplicities $m_1,..m_k, m_1+..m_k=M$
Working on small mutually disjoint discs around each $z_q$, one gets by Rouche that there is an $n_q$ st $f_n$ has $m_q$ zeroes close to $z_q$ for all $n \ge n_q, q=1,...k$. Picking $n(M) \ge n_q$ for all $q$ we get that $f_n$ has at least $M$ zeroes for $n \ge n(M)$ so by hypothesis we have $M \le N$ and we are done)
Note that an easy example like $f_n(z)=z-1+1/n \to z-1$ shows that $f_n$ has each a zero in the open unit disc, but $f$ doesn't so the inequality can be strict
