If $\Omega$ is a bounded open subset of $\mathbb{R}^n$, $\mathcal{B}_\Omega$ is the set of Borel subsets of $\Omega$, $\mu$ is the Lebesgue measure of $\mathbb{R}^n$, $H^1_0(\Omega)$ is the closure of $C^\infty_c(\Omega)$ in the Sobolev space $H^1(\Omega)$ with respect to the Sobolev norm and $V$ is a finite dimensional subspace of $H^1_0(\Omega)$ whose elements are continuous function, is it true that there exists $\delta>0$ such that $$\lim_{\|u\|\rightarrow\infty, u\in V}\left(\sup_{\Omega_0\in\ \mathcal{B}_\Omega, \mu(\Omega_0)>\delta}\left(\inf_{x\in\Omega_0}|u(x)|\right)\right)=+\infty?,$$ i.e. $$\exists \delta >0, \forall M>0, \exists R>0, \forall u\in V, (\|u\|>R)\implies \left(\exists\Omega_0\in\mathcal{B}_\Omega,(\mu(\Omega_0)>\delta)\land(\inf_{x\in\Omega_0}|u(x)|>M)\right)?$$ Or, sequentially: is it true that for each finite dimensional vector subspace of $H^1_0(\Omega)$ made of continuous functions, there exists $\delta>0$ such that for all $M>0$ there exists $R>0$ such that for each sequence $(u_n)_{n\in\mathbb{N}}\subset V$ diverging in $H^1_0(\Omega)$, if $N\in\mathbb{N}$ is such that $\forall n \ge N, \|u_n\|\ge R$ then there exists a sequence $(\Omega_n)_{n\ge N}\subset\mathcal{B}_\Omega$ such that $\forall n\in\mathbb{N}, \mu(\Omega_n)>\delta$ and $\forall n\in\mathbb{N}, \forall x\in\Omega_0, |u_n(x)|\ge M$?
Obviously the result is false if we remove the finite dimensional condition, as we can see taking a sequence of bounded continuous functions oscillating more and more... however I think that this should be the only obstruction to get the result and that somehow the finite dimensional condition should rule out this phenomenon, but I'm having trouble proving it...