Let $f_k\in L^\infty[0,1]$ and $g_k\in L^2[0,1]$ be two sequences such that $f_k\to f$ a.e., $\left\|f_k\right\|_{L^\infty}\leq c$ for any $k$ and $g_k\rightarrow g$ weakly in $L^2[0,1]$. Why does $f_k g_k\rightarrow fg$ weakly in $L^2[0,1]$?
We have to show $$ \int_0^1 f_kg_k\phi dx\rightarrow \int_0^1fg\phi dx $$for any $\phi$ in the dual of $L^2[0,1]$, so $\phi\in L^2[0,1]$. We have \begin{align*} \left|\int_0^1 f_kg_k \phi-fg\phi dx\right|&=\left|\int_0^1 f_kg_g\phi-f_kg\phi+f_kg\phi-fg\phi dx \right|\\ &\leq\|f_k\|_{L^\infty}\int_0^1|g_k\phi-g\phi|+\int_0^1|f_kg\phi-fg\phi|. \end{align*}How can you go on and show $f_kg_k\rightarrow fg$ weakly in $L^2[0,1]$?
There may well be a simpler or more elegant solution. We may as well simplify the notation by assuming $||f_k||_\infty\le 1$; since $||g_k||_2$ is bounded let's assume $||g_k||_2\le 1$ as well.
Say $\epsilon>0$. Choose $\delta>0$ so $m(E)<\delta$ implies $$\left(\int_E|\phi|^2\right)^{1/2}<\epsilon.$$
Egoroff's Theorem shows that there exists $F$ so that $f_k\to f$ uniformly on $F$ and $m(E)<\delta$ if $E=[0,1]\setminus F$. So for every $k$ we have $$\int_E|f_kg_k\phi|\le\epsilon.$$If $k$ is large enough then $|f_k-f|<\epsilon$ on $F$, so $$\int_F|(f_k-f)g_k\phi|\le\epsilon||\phi||_2.$$And $$\int_Ffg_k\phi=\int g_k(f\phi\chi_F)\to\int g(f\phi\chi_F)=\int_F fg\phi.$$