If $U$ is connected, any two sections $U \to \mathfrak S$ either coincide or have disjoint images (Is my proof correct?) I tried proving the following statement by Ahlfors, page 287:

If $U$ is connected and $\varphi,\psi: U \to \Gamma(U,\mathfrak S)$, then either $\varphi$ and $\psi$ are identical, or the images $\varphi(U)$ and $\psi(U)$ are disjoint. Indeed, the sets with $\varphi - \psi = 0$ and $\varphi- \psi \neq 0$ are both open.

Here $\mathfrak S$ is the sheaf of germs of analytic functions over some open set $D \subseteq \mathbb C$, and $\Gamma(U,\mathfrak S)$ is the set of all sections from an open set $U \subseteq D$.

Mainly, what I tried is expanding on the openness of the sets $$A=\{ \zeta \in U: \varphi(\zeta)-\psi(\zeta)=\mathbf{0}_\zeta \} \\
B=\{ \zeta \in U: \varphi(\zeta)-\psi(\zeta) \neq \mathbf{0}_\zeta \} $$
$A$ can be viewed as the inverse image $$(\varphi-\psi)^{-1}[\omega(U)] $$ where $\omega:\zeta \mapsto \mathbf{0}_\zeta$ is the zero section. Since it is an open map, and $\varphi-\psi$ is continuous it follows that $A$ is open.
However, proving that $B$ is open turned out to be more problematic for me. I couldn't do it using representation as an inverse image, so I tried the more direct approach: Let $\zeta_0 \in B$ and suppose that in every arbitrarily small disk $\Delta(\zeta_0,r)$ there exists a point $\zeta_r$ such that $(\varphi-\psi)(\zeta_r)=\mathbf{0}_{\zeta_r} \equiv \omega(\zeta_r)$. Thus we may extract a sequence $\{\zeta_n \}_{n=1}^\infty$ which tends to $\zeta_0$, such that for all $n$, $(\varphi-\psi)(\zeta_n)=\omega(\zeta_n)$. Since both $\varphi-\psi$ and $\omega$ are continuous, they are sequentially continuous, and taking the limit as $n \to \infty$ yields the contradiction $\varphi(z_0)-\psi(\zeta_0)=\mathbf{0}_{\zeta_0}$. It follows that $B$ is open as well.
Lastly, $U=A \coprod B$, and from connectedness either $U=A$ or $U=B$. In the former $\varphi,\psi$ coincide, and in the latter they have disjoint images (this is because if they share a value, they must share it at the same point).
Is this all correct? If not, please help me correct it. Thanks!
 A: Minor point: "$\varphi,\psi: U \to \Gamma(U,\mathfrak S)$" should be either  $\varphi,\psi  \in \Gamma(U,\mathfrak S)$   or $\varphi,\psi: U \to  \mathfrak S $, but not the mix of both notation. 
Your proof of the openness of $A$ is correct.  
Before discussing the openness of $B$, let's note that Ahlfors placed the quoted statement incorrectly. It comes    before he restricts consideration to the sheaf of analytic germs. But for general sheaves the quoted statement is false. 
For example, consider the sheaf of germs of continuous functions on $\mathbb R$. Define $f,g\in C(\mathbb R)$ by  $f(x)\equiv 0$ and $g(x)=  \max(x,0) $. Each continuous functions induce  a section of the sheaf, because for each $x$ it 
 determines a germ at $x$. Let $\varphi$ and $\psi$ be the  sections induced by $f$ and $g$ respectively. The set 
 $\{\varphi=\psi\}=(-\infty,0)$ is open, but its complement $\{\varphi\ne \psi\}=[0,\infty) $ is not. 
Your argument for the openness of $B$ implicitly assumes that the limit of a sequence is unique. But sheaves can be
non-Hausdorff, e.g.,  the sheaf of the germs of continuous functions. Indeed, in the preceding example $\varphi(0)\ne \psi(0)$, 
but both these points are limits of the sequence $\varphi(-1/n)$. 
To fix the proof, you should use the fact that you are dealing with analytic germs. For example: let $\zeta\in B$. Pick some function elements $(f,\Omega)$ and $(g,\Omega)$ that determine the germs $\varphi(\zeta)$ and $\psi(\zeta)$,
respectively. Since the zeros of $f-g$ are isolated, by shrinking $\Omega$ we may assume $f\ne g$ in $\Omega\setminus \{\zeta\}$. 
Hence, at each $w\in \Omega$ the germ  $(f-g)_w$ is nonzero. The set of all germs determined by $(f-g,\Omega)$ is an open subset of 
$\mathfrak S$, hence its preimage under $\varphi-\psi$ is an open subset of $U$. This subset contains $\zeta$, and is contained in $B$.  
