Given $k\in\mathbb{N}$, let $X_k=\{x\in[0,1]:\sum_{n=1}^\infty|u_n(x)|^2\le k\}.$ To prove your conclusion, it suffices to show that for every $k\in\mathbb{N}$, $m(X_k)=0$. To prove $m(E_k)=0$, it suffices to show that if $E$ is a measurable subset of $X_k$ with $km(E)<1$, then $m(E)=0$.
Fixing such a set $E$, since $\{u_n\}_{n=1}^\infty$ is an orthonormal basis of $L^2[0,1]$, $\sum_{n=1}^\infty\int_{E}\bar{u}_ndm\cdot u_n$ converges to $\chi_E$ in $L^2[0,1]$. Therefore, $\sum_{n=1}^\infty\int_{E}\bar{u}_ndm\cdot u_n\chi_E$ also converges to $\chi_E$ in $L^2[0,1]$. Since $E\subset X_k$, by Cauchy-Schwarz inequality, for every $x,y\in E$, $\sum_{n=1}^\infty|u_n(x)\bar{u}_n(y)|\le k$. It follows that
$\chi_E=|\sum_{n=1}^\infty\int_{E}\bar{u}_ndm\cdot u_n\chi_E|\le km(E)\chi_E<\chi_E \quad\mbox{a.e. on } E,$ which implies that $m(E)=0$.