Let $f$ be a holomorphic function on $P=U(z_o,r) \setminus \{z_o\}$ with an essential singularity in $z_o$. I would like to show that there is some $k \in \mathbb{C}$ such that : for every $ \epsilon$, there exists an $x$ in $P \cap U(z_o,\epsilon)$ for which $f(x) = k$.
I only know the Casorati–Weierstrass theorem about $f(P \cap U(z_o,\epsilon))$ being dense for all $\epsilon$, and would like to avoid using Picard's great theorem.
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An approach that might work (just an outline, not rigorous, and it's probably unnecessarily complicated), which I would be happy if someone checked, and told me whether it's going anywhere:
By the Casorati–Weierstrass theorem we know that $f(P \cap U(z_o,\epsilon))$ is dense for all $\epsilon$. Take some sequence of decreasing punctured neightborhoods $P_i$ around $z_0$, then $f(P_i)$ is dense for all $i$.
The following fact is true - intersection of a sequence of dense, open subsets, is still dense - this statement is equivalent to $\mathbb{C}$ being a a Baire space, but $\mathbb{R}^2$ is complete, and thus a Baire space. If we show, that the image of $P_i$ is open (for now, let us presume it's true), then we will know that the intersection of $f(P_i)$ is dense, and thus non-empty, so we can take a $k$ inside of it.
For an arbitrarily small punctured neightboorhood $P$, we can take some $P_i$ inside of $P$, and we know that $k \in f(P_i)$, so $f(x)=k$ for some $x \in P$, finishing the proof.
We will now show that $f(P_i)$ is open. Take some $\beta \in f(P_i)$, and consider it's pre-image $\alpha$. We can take an neightboorhood $U \subseteq P_i$, $\alpha \in U$. We know that $f$ is holomorphic and non constant on $P_i$ - because of this, it cannot be constant on $U$. This follows from this statement - if $\{ x : f(x)=g(x) \}$ has an accumulation point in an open, connected set $A$, and $f$ and $g$ are holomorphic, then $f=g$ on $A$.
Take $\tilde{f} = f - \beta$. Take it's series expansion $\sum_{n=1}^\infty a_n(z-\alpha)^n$. We know $\tilde{f}$ is non-constant, so some $a_i$ is non-zero. For simplicity, presume it's $a_1$. We can check that for a close enough neightborhood $V$ of $\alpha$, $\tilde{f}$ behaves similarly to $a_1(z-\alpha)$, which shows that we can find an open subset $L$ of $\beta$ that is contained in image of $f(V) \subseteq f(P_i)$. Alternatively (and this might be an unavoidable step anyway), we can simply aim to check that $\tilde{f}$ acts as a homeomorphism/diffeomorphism on some neightborhood of $\alpha$.
For an arbitrary $\beta \in f(P_i)$ we have found an open set $L$, $\beta \in L$, that is contained in $f(P_i)$, showing that $f(P_i)$ is open, finishing the proof.