Fix $\alpha>0$ and let $U=B_1(0)$. Show that if $u\in H^1(U)$ and $|\{x\in U:u(x)=0\}|\ge \alpha$, then there exists a constant $C=C(n, \alpha)>0$, such that $$\int_U u^2dx \le C\int_U {|Du|}^2dx.$$
I tried to prove by contradiction. Suppose there exists a sequence $u_n \in H^1(U)$ with $|\{x \in U: u^n(x)=0\}|\ge \alpha$ such that $\int_U u_n^2dx \ge n\int_U {|Du_n|}^2dx$. However I was stuck here to obtain a contradiction.
Suppose there is a sequence $u_n$ such that $\int_U u_n^2dx \ge n\int_U {|Du_n|}^2dx$. Normalize it so that $\int_U u_n^2dx =1$. Then $\int_U {|Du_n|}^2dx \to 0$. Since the sequence $\{u_n\}$ is bounded in $H^1(U)$, it has a weakly convergent subsequence, which is again denoted $\{u_n\}$. Let $u$ be its limit. The Dirichlet functional $\int_U {|Du|}^2dx$ is convex and therefore lower semicontinuous under weak convergence. This implies $$ \int_U {|Du|}^2dx \le \liminf_{n\to\infty} \int_U {|Du_n|}^2dx = 0 \tag1$$ hence $u$ is constant.
Weak convergence in $H^1(U)$ implies strong convergence in $L^2(U)$, hence $u_n\to u$ in $L^2(U)$. But this is impossible. Indeed, if $u\equiv c$ with $c\ne 0$, then for every $n$ we have $$ \int_U |u-u_n|^2 \ge c^2 |\{x\in U:u_n(x)=0\}| \ge c^2\alpha $$ And if $c=0$, then $$ \int_U |u-u_n|^2 = \int_U u_n^2 = 1 $$ by the normalization.
It is also possible to reduce the problem to the better-known version of Poincare's inequality $$ \int_U |u - \bar{u}|^2 \le C(U)^2 \int_U |\nabla u|^2, $$ where $\bar{u}$ is the average of $u$ over $U$. This way we can obtain a constant that depends on $\alpha$ explicitly.
Let $Z = \{ u = 0 \}$ be the zero set. Using Holder's inequality on $U \setminus Z$, we get \begin{align*} \int_U \bar{u}^2 & = |U| \cdot \bar{u}^2 \\ & = \frac{1}{|U|} \left( \int_{U \setminus Z} u \right)^2 \\ & \le \frac{|U \setminus Z|}{|U|} \int_{U \setminus Z} u^2. \end{align*} It is important that the coefficient above is strictly smaller than $1$, since this enables us to estimate $$ \| u \|_{L^2} \le \| u-\bar{u} \|_{L^2} + \| \bar{u} \|_{L^2} \le \| u-\bar{u} \|_{L^2} + \sqrt{1 - \frac{\alpha}{|U|}} \| u \|_{L^2}, $$ $$ \left( 1 - \sqrt{1 - \frac{\alpha}{|U|}} \right) \| u \|_{L^2} \le \| u-\bar{u} \|_{L^2} \le C(U) \| \nabla u \|_{L^2} $$
