Question: I am working on the following complex analysis problem, but I have 0 idea on how to do it. I would appreciate any help with it. 

Let $f(z)$ be a complex-valued $C^{\infty}$ function defined on a connected open $\Omega$ of the complex plane. Assume that $f(z)$ and $f^2(z)$ are both harmonic (i.e. the real and imaginary parts of these functions are harmonic). Prove that either $f(z)$ or $\overline{f(z)}$ is analytc in $\Omega$.

I have absolutely no idea on how to attack this problem, so any help with it would be highly appreciated! Thanks!

Answer: 
Expanding on Wojowu's comment, it is far easier to show that $\mathbb{R}^{2}$ less any countable set is path connected. From this connectedness is of course implied.

The argument ultimately comes down to the fact that the set of lines in the plane containing a particular point is uncountable. 

Let $A\subseteq\mathbb{R}^2$ be a countable set and let $x,y\in\mathbb{R}^{2}\setminus A$. Let $\mathcal{L}\_{x}$ be all lines in $\mathbb{R}^{2}$ that contain $x$. Then define $f:A\rightarrow\mathcal{L}\_{x}$ by defining $f(z)$ to be the unique line $L\in\mathcal{L}\_{x}$ that contains $z$. Then, because $A$ is countable and $\mathcal{L}\_{x}$ is uncountable we have that $\mathcal{L}\_{x}\setminus f(A)$ is uncountable. We can similarly define a set $\mathcal{L}\_{y}$ of lines containing $y$ and a function $g:A\rightarrow\mathcal{L}\_{y}$. Because $\mathcal{L}\_{x}\setminus f(A)$ and $\mathcal{L}\_{y}\setminus g(A)$ are uncountable we can choose nonparallel lines $L\_{1}\in\mathcal{L}\_{x}\setminus f(A)$ and $L\_{2}\in\mathcal{L}\_{y}\setminus g(A)$ which necessarily intersect. A path from $x$ to $y$ (or $y$ to $x$) in $\mathbb{R}^{2}\setminus A$ can be constructed using these lines.
For feedback:

Consider any two points $p = (x\_1,y\_1)$ and $q = (x\_2,y\_2).$

It is the case that $x\_1 \in \mathbb{I}$ is irrational or $y\_1 \in \mathbb{I}$ is irrational, and that $x\_2 \in \mathbb{I}$ is irrational or $y\_1 \in \mathbb{I}$ is irrational. Suppose without loss of generality that $x\_1$ is irrational. I will now consider the two cases: the case in which $x\_2$ is irrational and the case in which $y\_2$ is irrational. Suppose $x\_2$ is irrational. Then, there exists the line $\mathcal{L\_1} \in \mathbb{R}^2 \setminus \mathbb{Q}^2$ connecting $(x\_1,y\_1)$ to $(x\_1,i\_1),$ where $i\_1 \in \mathbb{I}.$ Moreover, there exists the line $\mathcal{L}\_2 \in \mathbb{R}^2 \setminus \mathbb{Q}^2$ connecting $(x\_1,i\_1)$ to $(x\_2,i\_1),$ where $i\_2 \in \mathbb{I}.$ And, there exists a line $\mathcal{L}\_3 \in \mathbb{R}^2 \setminus \mathbb{Q}^2$ connecting $(x\_2,i\_2)$ to $(x\_2,y\_2).$ So, the path $P = \mathcal{L}\_1 \cup \mathcal{L}\_2 \cup \mathcal{L}\_3 \in \mathbb{R}^2 \setminus \mathbb{Q}^2$ is a path connecting $(x\_1,y\_1)$ to $(x\_2,y\_2).$ The proof is almost identical when considering the case that $y\_2$ is irrational.

Is a proof like this even acceptable?