It is well know that a Brownian motion $(W(t))_t$ has independent increments. More precisely, given any $0\leq t<s,$ the increment $W(t)-W(s)$ is independent of the $\sigma$-algebra $\mathscr F(t)$ generated up till time $t$.
Then I believe that given any $0\leq t<u_1<u_2,$ the increment $W(u_2) -W(u_1)$ is also independent of $\mathscr F(t).$
Reason is as follows:
Since $W(u_2) - W(u_1) =(W(u_2 )-W(t))-((W(u_1 )-W(t))) $ and both $(W(u_2 )-W(t))$ and $((W(u_1 )-W(t)))$ are independent of $F(t)$. As difference between independent events is still independent, so $W(u_2) - W(u_1)$ is independent from $\mathscr{F}(t).$
I am not very sure the validity of the bold statement above.
If $(X,Y)$ is independent of $\mathcal{G}$, where $\mathcal{G}$ is a $\sigma$-algebra, then $\varphi(X,Y)$ is independent of $\mathcal{G}$ for any measurable function $\varphi$.
In your case, use the fact that $W_{u_2}-W_{u_1}$ is independent of $\mathcal{F}_{u_1}$ and $\mathcal{F}_t\subset \mathcal{F}_{u_1}$.