This is the problem 15 of Evans' PDE book.
That is, let $u \in H^1(\Omega) = W^{1,2}(\Omega)$, and $U = B(0,1)$ the unit ball in $\mathbb{R}^n$. Show there is a constant $C(\alpha, n)$ such that
$$ \int_U u^2 dx \le C \int_U |Du|^2 dx $$ provided that $$ |\{x \in U: u(x)=0\}|\ge \alpha. $$
On
Assume that $u$ is $\mathcal C^1$ (which should be enough by approximation).
Set $Z:=\{ z\in U;\; u(z)=0\}$, so that $\vert Z\vert\geq\alpha$.
For any $z\in Z$, we may write $$\int_U u^2(x)dx=\int_U (u(x)-u(z))^2dx=\int_U \left( \int_0^1 Du(z+t(x-z))\cdot (x-z)\, dt\right)^2dx\, . $$
Integrating with respect to $z\in Z$ and using Cauchy-Schwarz+Fubini, this gives
\begin{align*} \alpha \int_U u(x)^2dx&\leq \int_Z \int_U \left( \int_0^1 Du(z+t(x-z))\cdot (x-z)\, dt\right)^2dxdz\\ &\leq \int_Z\int_U \int_0^1 \vert Du(z+t(x-z))\vert^2 \vert x-z\vert^2 dtdxdz\\ &\leq 4\, \int_Z \int_0^1 \int_U \vert Du(z+t(x-z))\vert^2dxdtdz, \end{align*} because $\vert x-z\vert\leq 2$ for any $x,z\in U$.
Now, for each fixed $(z,t)\in Z\times (0,1)$, we can apply the change of variable formula, setting $y:=z+t(x-z)$. This change of variable maps $U$ onto the ball $B((1-t)z, t)$ with center $(1-t)z$ and radius $t$; and its jacobian is equal to $t^n$. So we get $$\int_U \vert Du(z+t(x-z))\vert^2dx\leq t^{-n} \int_{B((1-t)z, t)} \vert Du(y)\vert^2 dy.$$
Integrating now with respect to $(z,t)$, and observing that $y\in B((1-t)z,t)$ is equivalent to $z\in B(\frac{y}{1-t}, \frac{t}{1-t})$, we obtain \begin{align*}\int_Z \int_0^1 \int_U \vert Du(z+t(x-z))\vert^2dxdtdz&\\ &=\int_U \vert Du(y)\vert^2\left(\int_0^1 t^{-n} \left\vert Z\cap B(\frac{y}{1-t}, \frac{t}{1-t})\right\vert dt\right)dy \end{align*}
Now observe that $\left\vert Z\cap B(\frac{y}{1-t}, \frac{t}{1-t})\right\vert \leq \left\vert B(\frac{y}{1-t}, \frac{t}{1-t})\right\vert \leq 2^nt^n$ if $t\leq1/2$, and that $\left\vert Z\cap B(\frac{y}{1-t}, \frac{t}{1-t})\right\vert \leq\vert Z\vert\leq \vert U\vert$ if $t\geq 1/2$. Inserting these inequalities in the previous formula, we get \begin{align*} \int_Z \int_0^1 \int_U \vert Du(z+t(x-z))\vert^2dxdtdz&\leq \int_U \vert Du(y)\vert^2\left(\int_0^{1/2} 2^n dt+\int_{1/2}^1 t^{-n}\vert U\vert dt\right)\\ &:=C_n\,\int_U \vert Du(y)\vert^2dy. \end{align*}
Altogether, this gives $$\int_U u(x)^2dy\leq \frac{4C_n}{\alpha}\, \int_U \vert Du(y)\vert^2dy.$$
You can also argue by contradiction, similarly to what it's usually done to prove the Poincaré inequality $$\int\limits_\Omega|u-u_\Omega|^p\leq C\int\limits_\Omega|Du|^p$$ using the Rellich-Kondrachov compact embedding.
Proof Suppose by contradiction there is a sequence $(u_k)\subset W^{1,2}(\Omega)$ such that $$\|Du_k\|^2_{L^2}\le \frac1k \|u_k\|^2_{L^2}\qquad (*)$$ and without loss of generality by multiplying by a constant assume $\|u_k\|_{L^2}=1$. In particular $(u_k)$ is bounded in $W^{1,2}(\Omega)$, thus a subsequence converges weakly to some $u\in W^{1,2}(\Omega)$ and by Rellich-Kondrachov $u_k\to u$ strongly in $L^2$. Moreover by $(*)$ $Du_k\to 0$ strongly in $L^2$, therefore $u$ must be constant (it would suffice $Du_k\rightharpoonup 0$ weakly in $L^2$). Indeed for any test function $\phi$ $$\int\limits_\Omega Du\,\phi=-\int\limits_\Omega u\, D\phi=\lim_{k\to\infty} -\int\limits_\Omega u_k\, D\phi=\lim_{k\to\infty}\int\limits_\Omega Du_k\,\phi=0.$$ Setting $N=\{x\in \Omega:u(x)=0\}$, we have that $|N|\ge \alpha$: indeed for every $\lambda >0$ $$\left|\left\{x:|u(x)|\ge\lambda\right\}\right|\leq \left|\left\{|u(x)-u_k(x)|>\frac{\lambda}{2}\right\}\right|+\left|\left\{|u_k(x)|>\frac{\lambda}{2}\right\}\right|$$ and the first term goes to zero as $k\to\infty$, while the second term is $\le |\Omega|-\alpha$ by hypothesis.
Now from the strong convergence in $L^2$ we infer $\|u\|_{L^2}=1$, while from the fact that $u$ is constant and vanishes on $N$ (which has positive measure) we deduce $u=0$, contradiction.
This approach has the disadvantage of not giving explicit bounds on the involved constants, but can be used to prove many different Poincaré-like inequalities when the only constant function that satisfies the "boundary" conditions (in this case $u=0$ on a positive measure set) is the zero function.