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: 
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).
The notion of a vector space is abstract and it can be applied to Functions (sometimes these spaces are called Function Spaces) and these spaces are the subject of Functional Analysis. 

It is important to get away from the geometric representation of a vector space - they are special cases and impossible to think about when we move into higher dimensional spaces. Instead, consider our definition of a vector space:

Let $V$ be a set which is closed under vector addition and multiplication of vectors by scalars, then we call V a vector space. i.e.

$$ \forall x, y \in V \quad \forall c,d \in R $$

$$ cx+dy \in V $$

When we are talking about function spaces we are talking about mappings from a set $X$ to a vector space (over a field but don't get too bugged down in this if you don't know what this means) $V$ (note that $X$ can also be a vector space, then we can consider the function space to be a linear mapping). 

A more simple way of thinking about them is considering them as a collection of functions that share characteristics of their range and whose codomains' are commonly the same.

Example time! 

The function space $C(\mathbb R^n)$ represents all functions who are continuous in $\mathbb R^n$. e.g. $f(x)=x \in C(\mathbb R^1 )$

A Hilbert Spaces are subspaces of $C(\mathbb R^n)$ and often referred to as generalisations of Euclidean space.