Even though $\mathbb R^2$ and $\mathbb C$ are isomorphic as real vector spaces, they are very different in some algebraic respects, which crucially influence the notion of differentiability. Recall the key idea that a function is differentiable at a point if it has a best linear approximation (more precisely, a constant plus a linear transformation) near that point.
In the context of functions $\mathbb R^2\to\mathbb R^2$, "linear transformation" means a transformation that respects addition of vectors and multiplication by real scalars. In other words, the transformation respects the real vector space structure of $\mathbb R^2$. It is well known that such linear transformations are given by $2\times2$ real matrices (once one has chosen a basis for $\mathbb R^2$).
In the context of functions $\mathbb C\to\mathbb C$, on the other hand, "linear transformation" means a transformation that respects addition of vectors and multiplication by complex scalars. In other words, the transformation respects the complex vector space structure of $\mathbb C$. It is well known that such linear transformations are just multiplication by a single complex number. That's much more restrictive than multiplying by an arbitrary $2\times 2$ real matrix.
Specifically, if we use $\{1,i\}$ as our basis for $\mathbb R^2$, then the $2\times 2$ real matrices that correspond to complex linear transformations are just those of the form $\pmatrix{a & b\\-b & a}$. Because complex linear transformations are a very special sort of real linear transformations, complex differentiable functions are a very special sort of real differentiable functions. That "specialness" ultimately accounts for all the miraculous consequences of complex differentiability.
Write
$$f(x,y)=u(x,y)+i v(x,y),\quad g(\xi,\eta)=a(\xi,\eta)+ib(\xi,\eta)$$
where $u$, $v$, $a$, $b$ are realvalued functions defined in $A$, resp. $B$. Then by definition of $g$ one has
$$a(\xi,\eta)+ib(\xi,\eta)=g(\xi+i\eta)=\overline{f(\xi-i\eta)}=u(\xi,-\eta)-iv(\xi,-\eta)$$
and therefore
$$a(\xi,\eta)=u(\xi,-\eta),\quad b(\xi,\eta)=-v(\xi,-\eta)\qquad\bigl((\xi,\eta)\in B\bigr)\ .$$
It follows that, e.g. $$b_\eta(\xi,\eta)=-v_y(\xi,-\eta)\cdot(-1)=v_y(\xi,-\eta)\ .$$
Since $u$ and $v$ satisfy the CR-equations in the variables $x$ and $y$ we conclude that
$$a_\xi(\xi,\eta)=u_x(\xi,-\eta)=v_y(\xi,-\eta)=b_\eta(\xi,\eta)\ ,$$
and similarly
$$a_\eta(\xi,\eta)=-u_y(\xi,-\eta)=v_x(\xi,-\eta)=-b_\xi(\xi,\eta)\ .$$
This shows that $g$ fulfills the CR-equations in the variables $\xi$ and $\eta$.
But there is also a direct approach, which in my view is simpler and more in tune with a complex world description.
As $f$ is holomorphic in $A$, for each point $z_0\in A$ (held fixed in the following) there is a complex number $C$ such that
$$f(z)-f(z_0)=C(z-z_0)+o(|z-z_0|)\qquad (z\in A, \ z\to z_0)\ .$$
Let a point $w_0\in B$ be given, and put $z_0:=\bar w_0$. Then by definition of $g$ one has
$$g(w)-g(w_0)=\overline{f(\bar w)}-\overline{f(\bar w_0)}=\overline{f(\bar w)-f( z_0)}=\overline{C(\bar w -z_0)+o(|\bar w-z_0|)}\qquad(w\in B)\ .$$
As $|\bar w -z_0|=|w-w_0|$ it follows that
$$g(w)-g(w_0)=\bar C(w-w_0)+o(|w-w_0|)\qquad(w\in B, \ w\to w_0)\ .$$
It follows that $g'(w_0)=\bar C$, and as $w_0\in B$ was arbitrary, we conclude that $g$ is holomorphic in $B$.
Best Answer
So you want to show that if $f(z)$ is holomorphic, then $\overline{f(\bar z)}$ is holomorphic too.
I think it will be easiest not to split into real and imaginary parts -- so no Cauchy-Riemann -- but instead work directly from the definition of differentiability.
A natural guess would be that $\frac{d}{dz} \overline{f(\bar z)}$ would be $\overline{f'(\bar z)}$. Can you show that this is in fact the case?
For the second part, perhaps show that if $g(z)$ and $\overline{g(z)}$ are both holomorphic, then $g$ is constant. (Here, using Cauchy-Riemann feels more promising).