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).
I'll take $A\subseteq\mathbb{C}$ to be an open set. A function $f:A\rightarrow
\mathbb{C}$ is holomorphic in $A$ if for every $z_{0}\in A$ there exists the
limit
$$
\lim_{z\rightarrow z_{0}}\frac{f(z)-f(z_{0})}{z-z_{0}}=\ell\in\mathbb{C},
$$
which just says that the derivative $f^{\prime}(z_{0})$ exists in $\mathbb{C}%
$. Some books unfortunately require $f^{\prime}$ to be continuous, which just
muddles the waters. It is not needed, you get it for free.
Once you know that $f$ is holomorphic, you can prove that $f^{\prime}$ is
continuous and it is itself holomorphic, and then you can prove that there
exist derivatives of any order and are all holomorphic. To prove this you fix
an open ball $B$ in $A$ and prove that for every rectifiable close curve
$\gamma$ contained in $B$ you have that $\int_{\gamma}f(z)\,dz=0$. Note that
to make sense of the integral you only need $f$ to be continuous and not even
differentiability. Cauchy proved this theorem assuming that $f^{\prime}$ was
continuous (and this is why some books add this to the definition) but later
Goursat proved that the result continues to hold assuming that $f$ is only differentiable.
Once you have $\int_{\gamma}f(z)\,dz=0$ for every rectifiable close curve
$\gamma$ contained in $B$ you prove Cauchy's formula, that is, that for every
$z_{0}\in B$
\begin{equation}
f(z_{0})=\frac{1}{2\pi i}\int_{\partial B(z_{0},r)}\frac{f(z)}{z-z_{0}%
}dz\label{cauchy formula}%
\end{equation}
where the closed ball $\overline{B(z_{0},r)}$ is contained in $B$. The point
now is that the right-hand side is a function $h(z_{0})$ which is given by an
integral depending on a parameter and as a function of $z_{0}$ you have that
$\frac{1}{z-z_{0}}$ is as regular as you want. So you can use theorems on
differentiation under the integral sign to conclude that that the right-hand
side has derivatives of any order and they are all continuous. In turn the
same is true for the left-hand side. So now you know that $f$ has derivatives
of any order and they are all continuous. At this point, if you define
$u(x,y):=$ real part of $f(x+iy)$ and $v(x,y):=$immaginary part of $f(x+iy)$,
then $u$ and $v$ are $C^{\infty}$ because they are given by the composition of
the function $(x,y)\mapsto x+iy$ which is $C^{\infty}$ with the function $f$
which is also $C^{\infty}$. In particular you can use the chain rule and take
second derivatives of $u$ and $v$ and prove that the Cauchy-Riemann equations
are satisfied and that $u$ and $v$ is harmonic (the mixed derivatives cancel
out because $u$ and $v$ are $C^{\infty}$). Hope this answers your second questions.
Now the converse implication is not perfect. There is a nice review article of
Gray and Morris
. One good result is the following. Take $U\subseteq
\mathbb{R}^{2}$ be an open set and let $u:U\rightarrow\mathbb{R}$ and
$v:U\rightarrow\mathbb{R}$ be such that there exist the partial derivatives
$\partial_{x}u$ and $\partial_{y}u$ and $\partial_{x}v$ and $\partial_{y}v$.
Note that the existence of partial derivatives DOES not imply that $u$ and $v$
are differentiable in $U$. You also assume that $u$ and $v$ satisfy the
Cauchy-Riemann equations.
These hypotheses are not enough, since they do not even imply that $u$ and $v$
are continuous. The function
$$
f(z):=\left\{
\begin{array}
[c]{ll}%
\exp(-z^{-4}) & \text{if }z\neq0\\
0 & \text{if }z=0,
\end{array}
\right.
$$
does not have a derivative at $z=0$ but the corresponding functions $u$ and
$v$ satisfy the Cauchy-Riemann equations in $\mathbb{R}^{2}$.
So you need an extra hypothesis. There are several variants. One that I like
is that $u:U\rightarrow\mathbb{R}$ and $v:U\rightarrow\mathbb{R}$ admit
$\partial_{x}u$ and $\partial_{y}u$ and $\partial_{x}v$ and $\partial_{y}v$ in
$U$, $u$ and $v$ satisfy the Cauchy-Riemann equations, and that the function
$f(z):=u(x,y)+iv(x,y)$, where $z=x+iy$, is continuous in the domain
$A=\{z=x+iy:\,(x,y)\in U\}$. The idea of the proof is to to mollify the
functions $u$ and $v$, taking $u_{\varepsilon}=\varphi_{\varepsilon}\ast u$
and $v_{\varepsilon}=\varphi_{\varepsilon}\ast v$, where $\varphi
_{\varepsilon}$ is a nice kernel, and prove that the $C^{\infty}$ functions
$u_{\varepsilon}$ and $v_{\varepsilon}$ satisfy the Cauchy-Riemann equations.
Using that, you can prove that $\int_{\gamma}f_{\varepsilon}(z)\,dz=0$ for
every rectifiable close curve $\gamma$ contained in $B$, where $f_{\varepsilon
}(z):=u_{\varepsilon}(x,y)+iv_{\varepsilon}(x,y)$, where $z=x+iy$, and then
continue as before to prove that $f_{\varepsilon}$ is analytic. In turn you
get the Cauchy formula (1) for $f_{\varepsilon}$ and then since $f$ is
continuous, you can let $\varepsilon\rightarrow0$ to prove that $f$ satisfies
the Cauchy formula and so is holomorphic.
Hope this answers your first question. Read the paper, it is well-written.
Best Answer
This is essentially the open mapping theorem, which states that a holomorphic, non-constant function is always an open map(i.e. it sends open subsets of its domain to open subsets of $\mathbb{C}$).
Now, the reals are a closed subset of $\mathbb{C}$, so if you have a holomorphic map $f: U \to \mathbb{R}$, where $U$ is open and connected, that map is constant; if it wasn't, $f(U)$ should be open in $\mathbb{C}$.
If we drop the assumption that $U$ is connected, then the map is constant at each component of $U$, but those values may be different for each component.