There exists a unique function $u\in C^0[-a,a]$ which satisfies this property The problem:

Let $a>0$ and let $g\in C^0([-a,a])$. Prove that there exists a unique function $u\in C^0([-a,a])$ such that $$u(x)=\frac x2u\left(\frac x2\right)+g(x),$$ for all $x\in[-a,a]$. 

My attempt At first sight I thought to approach this problem as a fixed point problem from $C^0([-a,a])$ to $C^0([-2a,2a])$, which are both Banach spaces if equipped with the maximum norm. However i needed to define a contraction, because as it stands it is not clear wether my operator $$(Tu)(x)=\frac x2u\left(\frac x2\right)+g(x)$$ is a contraction or not. Therefore I tried to slightly modify the operator and I picked a $c>a>0$ and defined $$T_cu=\frac 1cTu.$$ $T_cu$ is in fact a contraction, hence by the contraction lemma i have for granted the existence and the uniqueness of a function $u_c\in C^0([-a,a])$, which is a fixed point for $T_cu.$ Clearly this is not what I wanted and it seems difficult to me to finish using this approach. Am I right, is all what I have done useless? And if this were the case, how to solve this problem?
Thanks in advance.

As noted already, your approach works for $a<2$.
Now suppose that $a\geq 2$. We will first construct $u$ on [-1,1] and then double the width of this interval iteratively. Given a function $g$, the equation
$$\tag{$\ast$} u(x) = \frac{x}2 u\left(\frac{x}2\right) + g(x)$$
uniquely determines a function $u_0$ on [-1,1]. Now define $u_1$ by
$$u_1(x)=\begin{cases} u_0(x) &\text{if $x\in[-1,1]$} \\ \frac{x}2 u_0\left(\frac{x}2\right) + g(x) &\text{otherwise}.\end{cases}$$
We now have a function $u_1$ on [-2,2] satisfying $(\ast)$. As long as $2^{k+1}<a$, continue this way defining
$$u_{k+1}(x) = \begin{cases} u_{k}(x) &\text{if $x\in[-2^{k},2^{k}]$} \\ \frac{x}2 u_{k}\left(\frac{x}2\right) + g(x) &\text{otherwise}.\end{cases}$$
At a certain point, $2^{k+1}\geq a$. Define $u_{k+1}$ only on $[-a,a]$, then stop. Note that all the $u_k$ are continuous.
It is clear that the last function you constructed is the unique solution to the problem (solution by construction, uniqueness because $u_0$ is unique and because $(\ast)$ forces all the other function values).
