Question: Let $f$ be an analytic function in an open set $U$. Let $V = \{z \in \mathbb{C} : \overline{z} \in U\}$. Define $g$ on $V$ by $g(z) = \overline{f(\overline{z})}$. Show that $g$ is analytic on $V$ without using the fact that $g$ is holomorphic.
Proof Attempt: Choose $z_0 \in V$. Then $\overline{z_0} \in U$. Since $f$ is analytic on $U$, there is an $r > 0$ and a sequence of complex numbers $\{a_n\}_{n = 0}^{\infty}$ such that for $\overline{z} \in D(\overline{z_0}, r)$, we have $$f(\overline{z}) = \sum_{n = 0}^{\infty} a_n(\overline{z} - \overline{z_0})^n$$ It follows that $$g(z) = \overline{f(\overline{z})} = \sum_{n = 0}^{\infty} \overline{a_n}(z - z_0)^n$$ for all $z \in D(z_0, r)$. Thus, $g$ is analytic on $V$.
Since conjugation is a homeomorphism, $V$ is open.
Let $c\in V$; then $\bar{c}\in U$, so there is an $r>0$ such that $D(\bar{c},r)\subseteq U$ and $f$ can be written as $$ f(z)=\sum_{n=0}^{\infty}a_n(z-\bar{c})^n \tag{1} $$ for every $z\in D(\bar{c},r)$. Therefore $$ f(\bar{z})=\sum_{n=0}^{\infty}a_n(\bar{z}-\bar{c})^n $$ for every $z\in\overline{D(\bar{c},r)}=D(c,r)$.
The convergence is absolute on $D(\bar{c},s)$, for every $0<s<r$, so it is not restrictive to assume the series $(1)$ is absolutely convergent on $D(\bar{c},r)$ (by using $r/2$ instead of $r$).
Then the series $$ \sum_{n=0}^{\infty}\overline{a_n}(z-c)^n $$ is absolutely convergent on $\overline{D(\bar{c},r)}=D(c,r)$, so it defines an analytic function over this disk. Since $$ \sum_{n=0}^{\infty}\overline{a_n}(z-c)^n= \overline{\sum_{n=0}^{\infty}a_n(\bar{z}-\bar{c})^n}= \overline{f(\bar{z})} $$ we are done.
Your idea is good, but the convergence of the final series should be mentioned.