I want to prove that the space of harmonic function on an open set $\Omega\subset R^N$ $(N\geq 2)$ is uncountablely infinite dimensional.
That is, I want to prove that
$$A:=\{u\in C^0(\Omega),\,\,u\text{ is harmonic }\}$$ has uncountablely infinite dimensions...
I actually have no idea how to get start. From what I learned in Functional analysis, we usually prove some Banach space is uncountablely infinite dimensional by using Zorn's lemma. But here I can not give a norm on $A$ since $u$ could be unbounded...
In the plane, harmonic functions are exactly the real parts of (complex) analytic functions, which can be represented by power series with an infinite radius of convergence. E.g., you can linearly embed the space $\ell^\infty(\mathbb{C})$ of bounded sequences of complex numbers into the space of harmonic functions in the plane by $(a_n)_{n=1}^\infty \mapsto \mathrm{Re} \left(\sum_{n=1}^\infty n^{-n}{a_n} z^n\right)$.
The space $\ell^\infty(\mathbb{C})$ does not have a countable basis which is probably easiest to see by showing that the set of sequences $(a^n)_{n=1}^\infty$ for $0<|a|<1$ is linearly independent.
If $N \ge 2$ and $\Omega \subset \mathbb{R}^N$ non-empty and open, then you can associate to every harmonic function $u(x,y)$ in the plane the harmonic function $u_\Omega(x_1,x_2, \ldots, x_N) := u(x_1, x_2)$. Since a harmonic function in the plane is uniquely determined by its values on any non-empty open set, this shows the claim in full generality.