Question: I need to prove that the following ODE do not have solution in Real numbers domain:
$$
\begin{cases}
\big(y'\big)^2=y-\lvert x\rvert\\[1.2ex]
y(x=0)=0
\end{cases}
$$
I found the complex solution but have trouble in explaining why there is no real solution.

Answer: 
$\mathbb R^D$ is the set of all functions $f:D \to \mathbb R.$ If we define an addition $f+g$ and a scalar multiplication $ \alpha f$ in this set by

$(f+g)(x)=f(x)+g(x)$ and $( \alpha f)(x)= \alpha f(x)$,

then $\mathbb R^D$ is a real vector space ( of functions).
I'll offer a point of view that gives some concrete examples.

As people have mentioned, the only thing necessary to have a "vector space" is the ability to add objects together, and multiply them by scalars (subject to some special rules). We have this for functions that share a common domain and codomain. If we consider **all** functions with codomain $\mathbb{R}$, and fixed codomain $D$, we get an infinite-dimensional vector space (unless $D$ is a finite set). This is what is usually called $\mathbb{R}^{D}$. 

Now, if we want a **finite** dimensional vector space, what we are looking for is a subspace of $\mathbb{R}^{D}$ that can be spanned by a finite set of functions. Here span is the normal linear algebra concept, where we are allowed to take linear combinations of the functions, e.g. the span of functions $f(x)$ and $g(x)$ would look like $\{ af(x) + bg(x) \ : \ a,b \in \mathbb{R}\}$.

Some examples: 

1. If we take the set of constant functions $f(x) = c$ for $c \in \mathbb{R}$, this is a 1-dimensional vector space of functions, because any such function is just $c$ times the function $f(x) = 1$.
2. If we take the set of polynomials of degree less than $n$, we get a vector space of dimensions $n+1$, for example the polynomials with degree less than 4 gives a 5-dimensional vector space with basis $\{1,x,x^{2}, x^{3}, x^{4}\}$.
3. If we take linear combinations of $\sin{x}$ and $\cos{x}$, we get a vector space of dimension 2 containing functions of the form $\{a\sin{x}+b\cos{x} \ : \ a,b \in \mathbb{R}\}$ (it can be shown that $\sin{x}$ and $\cos{x}$ are not scalar multiples of each other, so are linearly independent).