Strengthened version of the Casorati-Weierstrass Theorem Suppose the $f$ has an essential singularity at $z=a$.Prove that if $c\in \mathbb{C}$,and $\varepsilon >0$ are given,then for each $\delta >0$ there is a number $b$,$|c-b|<\varepsilon$,such the $f(z)=b$ has infinitely many solution in $B(a;\delta)$.
This is an exercise from 'functions of one complex variable'.
I solved this question by using open mapping theorem and Baire Category Theorem to argue that $\bigcap_{i=1}^{\infty}f(\{z:0<|z-a|<1/n\})$ is a dense set.
Is there a solution which does not use Baire Category Theorem? Thank you.
 A: Here is an argument which avoids Baire, at least explicitly, though it feels very similar to Baire.
Fix $\epsilon, \delta>0$ and $c \in \mathbb C$. Consider the sequence of sets $W_n=f(\{z \in \mathbb C | 0<|z-a|<\frac 1 {n+N}\}$ for $n \in \mathbb Z^+$ where $N$ is such that $\frac 1 {n+N} >\delta$ for all $n$. By the original Casorati-Weierstrass each $W_n$ is dense in $\mathbb C$ and by the open mapping theorem each is open.
Then $W_1 \cap B_\epsilon (c)$ is non-empty and open so that it contains a point $x_1$ and the closed ball $\overline B_{r_1}(x_1)$ such that $r_1 \in (0,\frac 1 2)$. Build by induction a sequence $(x_n)_{n \in \mathbb Z^+}$ and $(r_n)_{n \in \mathbb Z^+}$ such that $r_n \in (0, 2^{-n})$ and $\overline B_{r_{n+1}}(x_{n+1}) \subset W_n \cap B_{r_n}(x_n)$. Then $(x_n)_{n \in \mathbb Z^+}$ is Cauchy and must converge to some $x$. We have that $x \in B_\epsilon(c) \cap \bigcap_{n \in \mathbb Z^+}W_n$ so it is within $\epsilon$ of $c$ and has infinitely many pre-images within $\delta$ of $a$, as desired.
