$f\in E \mapsto \underset{{x\in[0,1]}}\sup f(x)$ continuous 
Let $E=(\mathcal{C}[0,1],\Vert f\Vert_{\infty})$ and $x\in [0,1].$ 
  Denote $\phi: f\in E \mapsto \underset{{x\in[0,1]}}\sup f(x)\in \Bbb{R}.$
I would like to prove that $\phi$ is continuous.

It means that $\forall\varepsilon>0\exists \eta>0\forall \tilde{f}\in E, \Vert f-\tilde{f} \Vert_{\infty}<\eta\implies\vert \phi(f)- \phi(\tilde{f})\vert<\varepsilon.$
$1)$ I have the following inequality $\vert \phi(f)-\phi(\tilde{f})\vert\le\Vert f\Vert_{\infty}+\Vert \tilde{f}\Vert_{\infty}$ but it's useless.
$2)$ I tried using sequential characterization but that does not seem like a better idea
 A: In general we have: $$\sup_{x\in[0,1]} (f(x) + g(x)) \le \sup_{x\in[0,1]} f(x) + \sup_{x\in[0,1]} g(x)$$
Let $\varepsilon > 0$ and $\|f - \tilde{f}\|_\infty < \varepsilon$.
$$\sup_{x\in[0,1]}f(x) = \sup_{x\in[0,1]}\left((f(x) - \tilde{f}(x)) + \tilde{f}(x)\right) \le \sup_{x\in[0,1]} (f(x) - \tilde{f}(x)) +\sup_{x\in[0,1]}\tilde{f}(x)$$
In particular we have
$$\sup_{x\in[0,1]}f(x) - \sup_{x\in[0,1]}\tilde{f}(x) \le \sup_{x\in[0,1]} (f(x) - \tilde{f}(x)) \le \sup_{x\in[0,1]} |f(x) - \tilde{f}(x)| =\|f - \tilde{f}\|_\infty$$
Similarly we obtain
$$\sup_{x\in[0,1]}\tilde{f}(x) - \sup_{x\in[0,1]}f(x) \le \sup_{x\in[0,1]} (\tilde{f}(x) - f(x)) \le \sup_{x\in[0,1]} |\tilde{f}(x) - f(x)| =\|f - \tilde{f}\|_\infty$$
Hence,
$$\left|\phi(f) - \phi(\tilde{f})\right| = \left|\sup_{x\in[0,1]}f(x) - \sup_{x\in[0,1]}\tilde{f}(x)\right| \le \|f - \tilde{f}\|_\infty < \varepsilon$$
So $\phi$ is uniformly continuous.
A: I write $||*|| $ instead of ||*||_{\infty}.
Let $c>0$. From $||f-g||<c$ it follows that $-c <f(x)-g(x)<c$ for all $x \in [0,1]$, hence
$f(x)<g(x)+c$ for all $x \in [0,1]$, 
which gives $f(x)< \phi(g)+c$ for all $x \in [0,1]$ and therefore $\phi(f) \le \phi(g)+c$.
Conclusion: $\phi(f)-\phi(g) \le c$.
In the same way we get  $\phi(g)-\phi(f)\le c$.
Thus we have shown: if  $||f-g||<c$ , then $|\phi(f)-\phi(g)|\le c$.
