Finding an ON basis of $L_2$ The set $\{f_n : n \in \mathbb{Z}\}$ with $f_n(x) = e^{2πinx}$ forms an orthonormal basis of the complex space $L_2([0,1])$. 
I understand why its ON but not why its a basis?

It is known that orthonormal system $\{f_n:n\in\mathbb{Z}\}$ is a basis if 
$$
\operatorname{cl}_{L_2}(\operatorname{span}(\{f_n:n\in\mathbb{Z}\}))=L_2([0,1])
$$
where $\operatorname{cl}_{L_2}$ means the closure in the $L_2$ norm.
Denote by $C_0([0,1])$ the space of continuous functions on $[0,1]$ which equals $0$ at points $0$ and $1$. It is known that for each $f\in C_0([0,1])$ the Feier sums of $f$ uniformly converges to $f$. This means that
$$
\operatorname{cl}_{C}(\operatorname{span}(\{f_n:n\in\mathbb{Z}\}))=C_0([0,1])
$$
where $\operatorname{cl}_{C}$ means the closure in the uniform norm. 
Since we always have inequality $\|f\|_{L_2([0,1])}\leq\|f\|_{C([0,1])}$, then 
$$
\operatorname{cl}_{L_2}(\operatorname{span}(\{f_n:n\in\mathbb{Z}\}))=C_0([0,1])
$$
It is remains to say that $C_0([0,1])$ is dence subspace of $L_2([0,1])$, i.e.
$$
\operatorname{cl}_{L_2}(C_0([0,1]))=L_2([0,1])
$$
then we obtain
$$
\operatorname{cl}_{L_2}(\operatorname{span}(\{f_n:n\in\mathbb{Z}\}))=
\operatorname{cl}_{L_2}(C_0([0,1]))=L_2([0,1])
$$
