$Lemma:$ Let $V$ and $U$ be open sets in $C$ and $F:V\to U$ a holomorphic function. If $u:U\to C$ is a harmonic function, then $u\circ F$ is harmonic on $V$.
$proof.$ The thrust of lemma is purely local, so we may assume that $U$ is an open disc.We let $G$ be a holomorphic function in $U$ whose real part is $u$.Let $H=G\circ F$ and not that $u\circ F$ is the real part of $H$.Hence $u\circ F$ is harmonic because $H$ is holomorphic.
I can give a computational proof of this lemma with Cauchy-Riemann equations.But I can't understand the proof above.For $U$ can be an arbitrary open set,why we can assume that $U$ is an open disc?
If $U$ is not an open disk, then you can decompose it into a union of open disks. That is, since $U$ is open, for each $x\in U$ you can find $r_x > 0$, so that the open disk $D_{r_x}(x)$ with radius $r$ and centered at $x$ is a subset of $U$, and then $$ U = \bigcup_{x \in U} D_{r_x}(x), $$ so you can apply the proof for each disk separately. In particular, you can assume without loss of generality that $U$ is an open disk.