Let $V$ be a Banach space, $V\supseteq N\supseteq N_{2\rho}\supseteq N_\rho \supseteq N_\delta$. Let for every $u\in V$, $\phi_u:[0,\infty[ \to V$ be a solution of the initial value problem $$\begin{cases}\frac{\partial}{\partial t}\phi_u(t) = e(\phi_u(t)) \\\phi_u(0) = u\end{cases}$$ where $e:V\to V$ is locally Lipschitz-continuous and uniformly bounded by $||e||\leq 2$. Let $u\notin N$. In the proof they say because of the uniform boundedness $||e||\leq 2$ and because $V\setminus N$ and $N_\delta$ are separated by the "annulus" $N_{2\rho}\setminus N_\rho$ with width $\rho$ we have that the Lebesgue-measure $$\lambda(\{t\mid\phi_u(t)\notin N_\delta\}) \geq \frac{\rho}{2}.$$ I think, I understand the idea. If there is a $\tilde t$ such that $\phi_u(\tilde t)\in N_\delta$, there are $t\in [0, \tilde t[$, such that $\phi_u(t)\in N_{2\rho}\setminus N_\rho$ and as $\phi_u$ is changing with velocity of $2$ max, the estimate above somehow has to hold. If there is not such a $\tilde t$, $\phi_u(t)\notin N_\delta$ anyways holds for every $t>0$.
I want to do this more formally but I have problems with this. Can someone help me there?
The deformation lemma and its proof I found in Michael Struwe's "Variational Methods", Vol. 34
This follows from the fundamental theorem of calculus for the Cauchy integral. Namely we have $$ \phi_u(t) = \phi_u(0) + \int_0^t e(\phi_u(s)) ds$$ Hence, we get $$ \Vert \phi_0(t) - u \Vert = \left\Vert \int_0^t e(\phi_u(s)) ds \right\Vert \leq \int_0^t \Vert e(\phi_u(s)) \Vert ds \leq 2t $$ The triangle inequality for the Cauchy integral can be check on step functions and extends then to continuous functions. Now you can apply your reasoning that $u\in V$ and $dist(V, N_\rho) \geq \rho$, we get that for $0\leq t < \rho/2$ we are staying outside of $N_\rho$.