Is the following proof of continuity correct? I want to prove that $\max_{t\in[0,1]}x(t)$ is a continuous functional in $C[0,1]$.
I have that $\left | t-s \right |\leq 1$, $\forall t,s \in [0,1]$: then
$$
0=\left | \max_{t\in[0,1]}x(t)-\max_{s\in[0,1]}x(s) \right |\leq \left | x(t)-x(s) \right |<\epsilon,
$$
and I assume that  $\left | x(t)-x(s) \right |<\epsilon$ because $x(t)\in C[0,1]$. 
Are my assumptions correct?
 A: Note: I assume below that $E=\mathcal C([0,1], \mathbb R)$ is endowed with the $\sup$ norm $\Vert \cdot \Vert_\infty$.
Your proof is not correct as to prove the continuity of
$$\begin{array}{l|rcl}
\phi : & \mathcal C([0,1], \mathbb R) & \longrightarrow & \mathbb R \\
    & x & \longmapsto & \max_{t \in [0,1]}  x(t)\end{array} $$
you need to consider the difference of two maps $x,y \in E$.
So take $x, y \in E$. For $t \in [0,1]$ we have
$$x(t) - y(t) \le \Vert x-y \Vert_\infty$$ hence
$$x(t) \le \Vert x-y \Vert_\infty + y(t)$$ and therefore
$$x(t) \le \Vert x-y \Vert_\infty + \max_{t \in [0,1]}  y(t) = \Vert x-y \Vert_\infty + \phi(y)$$ which implies
$$\phi(x) = \max_{t \in [0,1]}  x(t) \le \Vert x-y \Vert_\infty + \max_{t \in [0,1]}  y(t) = \Vert x-y \Vert_\infty + \phi(y)$$ i.e.
$$\phi(x) - \phi(y) \le \Vert x-y \Vert_\infty$$
As this proof is symmetric in $x,y$ we also get
$$\phi(y) - \phi(x) \le \Vert x-y \Vert_\infty$$ and finally
$$\vert \phi(y) - \phi(x) \vert \le \Vert x-y \Vert_\infty$$
From there it is clear that $\phi$ is continuous as it is Lipschitz.
