Suppose $U=\mathbb{C}\setminus K$ for some compact subset $K$ of $\mathbb{C}$, and let $f:U\to\mathbb{C}$ be holomorphic on $U$.
Question. Under what conditions can we find some $\alpha\in\mathbb{C}$ such that $f(u)=\alpha$ for infinitely many $u\in U$? In other words, I am trying to find any level set of infinite cardinality.
Discussion.
If $K$ is a set of isolated points then $f$ is meromorphic on $\mathbb{C}$ and we get an $\alpha$ via the meromorphic version of Picard's Great Theorem.
On the other hand, an $\alpha$ does not always exist. For instance, if $f(u)=u$ then obviously there is no such $\alpha$. To eliminate this kind of trivial counter-example, we could require that $f$ be a non-polynomial.
What additional conditions could we impose on $K$ and $f$ to guarantee the existence of a level set $f^{-1}(\alpha)$ of infinite cardinality? Perhaps it is enough for $f$ to have no analytic continuation beyond $U$.
Motivation.
This question is actually motivated by a problem in functional analysis. I have a Banach space $X$, a continuous linear operator $T\in\mathcal{L}(X)$, and a vector $x\in X$ such that the set $(T^nx)_{n=0}^\infty$ is linearly independent. I need a nonzero functional $x^*\in X^*$ such that the holomorphic map $u\in\mathbb{C}\setminus\sigma(T)\mapsto x^*((u-T)^{-1}x)$ admits a level set of infinite cardinality.
Thanks!
If $f$ has an essential singularity at $\infty$, then the existence of such an $\alpha$ will follow now from Picard's Great Theorem again.
In general however, we have counter-examples. Construct an injective function on $U=\mathbb{C} \setminus \overline { \mathbb{D}}$ as follows: Since $\dfrac{1}{z}$ maps $U$ to $\mathbb{D}\setminus \{0\}$ injectively, you can now compose it with your favourite random injective function on $\mathbb{D}\setminus \{0\}$.
In the non-compact case, let $U=\mathbb{C} \setminus (-\infty,0]$. $f(z)=\ln(z)$ defines a non-polynomial function on $U$ which is $1$-$1$, thus admitting no level set of infinite cardinality.