Show that exists $u_m$ such that $\phi(u_m) = \max\{\phi(u),u\in E\}$ 
Let $E = \{u:[0,1]\to \mathbb R | u(0)=0, lip(u)\leq 1\}$. Define $\phi(u) = \int_0^1 (u^2(x)-u(x))dx$
Show that exists $u_m\in E$ such that $\phi(u_m) = \max\{\phi(u),u\in E\}$

I intended to show that $\{\phi(u),u\in E\}$ is compact, in order to find that $\sup \{\phi(u),u \in E\} = \max\{\phi(u), u \in E\}$.
I can think of two ways of proving this, showing that $\phi$ is continuous and $E$ is a compact set, or showing that $\{\phi(u), u \in E\}$ is closed and limited.
I'm trying to use the second Idea:

Let $u \in E \implies |u(x)-u(y)|\leq |x-y|\; \forall x,y \in [0,1]$. As $u(0) = 0$, $|u(x)|\leq |x| = x\; \forall x \in [0,1]$.
So $\int_0^1((u^2(x)-u(x))dx \leq \int_0^1(|x|^2-|x|)dx  = -\frac{1}{6}$, therefore $\{\phi(u),u\in E\}$ is limited.
Now, how can I show it is closed? I don't believe that showing it's complement is open is a good idea and I can't think of a function to which this set would be the pre-image of something closed.
Any tips on how to continue OR alternative ways of showing that this $u_m$ exists?

This is an easy consequence of Arzela -Ascoli Theorem. The set $E$ is equi-continuous and bounded at $0$ which implies it is uniformly bounded. Hence any sequence $\{u_n\} \subset E$ has  a subsequence which converges uniformly. Can you now prove that $\{\phi (u):u\in E\}$ is closed?
