I want to show that $$\beta = \{1,\cos x, \sin x, \ldots, \cos nx, \sin nx\,,\ldots\}$$ is a basis for $L^2([-L,L]).$
I can see that $\beta$ is an othogonal set of functions (vectors) in $L^2([-L,L]).$ So it is a linearly independent set.
I don't understand why we can conclude that $\beta$ spans the space. In a finite dimensional space, if $S$ is a linearly independent subset of a vector space $V,$ with n vectors, and $dim(V)=n,$ then $S$ is a basis for $V.$ But in this infinite dimensional case, this clearly does not apply. More over, $\{\sin x, \sin 2x, ... , \sin nx, ...\}$ is an infinite linearly independent subset of $\beta$, and it is not a basis of $L^2([-L,L]).$
$\beta$ doesn't span $L^2=L^2[-L,L]$ as a vector space. Rather, if $X$ is the span of $\beta$ then $X$ is dense in $L^2$, that is $L^2=\overline X$, the closure of $X$ in $L^2$. This is the completeness of the trigonometric function, a standard but non-trivial result in Fourier theory.