FALSE: a holomorphic function with zero derivative on an arbitrary open set is constant

Let $z=x+iy\in U$ and choose $r:=\mathrm{\mid }x\mathrm{\mid }\mathrm{/}2>0$. If $\mathrm{\mid }w-z\mathrm{\mid }<r$, then $\mathrm{\mid }\mathrm{Re}w-x\mathrm{\mid }\le \mathrm{\mid }w-z\mathrm{\mid }<\mathrm{\mid }x\mathrm{\mid }\mathrm{/}2$, so $\mathrm{Re}w$ has the same sign as $x$. Thus $B(z,r)\subseteq U$, and [L1] shows that $U$ is open.

But $-1,1\in U$ and $f(-1)=0\mathrm{\ne }1=f(1)$, so $f$ is not constant on $U$.

The same ball $B(z,r)$ lies wholly in one half-plane, so $f$ is constant on it. For every sufficiently small nonzero $h$ with $z+h\in U$, the difference quotient $(f(z+h)-f(z))\mathrm{/}h$ is therefore $0$. Hence $f^{\mathrm{\prime }}(z)=0$.

Since $z\in U$ was arbitrary, [L2] says $f$ is holomorphic on $U$ and has derivative zero everywhere there.

Steps 3.1 and 1.2 refute the claim. The missing hypothesis is connectedness: $U$ is the disjoint union of two nonempty open half-planes, whereas [L3] requires a domain to be connected.