In this article we prove the little Picard theorem assuming the existence of a nonconstant holomorphic function $\lambda : \mathbb{C} \backslash \{0, 1\} \to \mathbb{C}$ which satisifies $\text{Re} \lambda(z) \leq 0$.
In the proof below we will repeatedly use the following fact: If $f$ is holomorphic on a domain (open and connected subset) $D$ of $\mathbb{C}$ and $K$ is a compact subset of $D$ such that $f(z) = 0$ for infinitely many $z \in K$, then $f \equiv 0$.
Theorem. (Little Picard theorem) If $f$ is a nonconstant entire function, then the image of $f$ must be entire complex plane minus at most one point.
Proof. Let $f$ be a nonconstant entire function. Assume that there exist two distinct points $\alpha$ and $\beta$ that do not lie in the range of $f$. We rescale $f$ by defining a function $g$ as
\[
g(z) = \frac{f(z) -\alpha}{\beta -\alpha}.
\] Note that $g$ is also an entire function and $0, 1 \not\in g(\mathbb{C})$. It now follows that $\lambda \circ g$ is a constant function since it is an entire function and its real part is bounded. This implies that
\[
(\lambda \circ g)'(z) = \lambda'(g(z)) g'(z) = 0 \tag{1}
\] for every $z \in \mathbb{C}$. Observe that $g(\overline{\mathbb{D}})$ is a compact subset of $\mathbb{C} \backslash \{0, 1\}$. If $g(\overline{\mathbb{D}})$ is finite, then $g$ must be a constant function so should be $f$, a contradiction. Now suppose that $g(\overline{\mathbb{D}})$ is infinite. Furthermore, if $g'(z) = 0$ for infinitely many $z \in \overline{\mathbb{D}}$, then $g’ \equiv 0$ and as a result $g$ must be constant. This is again a contradiction as $f$ is nonconstant. We now assume that $g'(z) = 0$ for only finitely many $z \in \overline{\mathbb{D}}$. Due to (1) it follows that $\lambda'(w) = 0$ for infinitely many $w \in g(\overline{\mathbb{D}})$. As a consequence $\lambda’ \equiv 0$ and hence $\lambda$ must be constant, a contradiction. Thus we conclude that our initial supposition about $\alpha$ and $\beta$ was false.