Prove that $\|A\|=1$ 
Consider $C^0([0,1])$ with the $d_0(f,g)=\int_0^1|f(x)-g(x)| dx$. Let be $A: C^0([0,1])\to C^0([0,1])$ defined by $A(x)(s)=sx(s)$. Show that $\|A\|=1$.

Proof: The first implication
$$\|A(x)\|_0=\|sx(s)\|_0=|s|\|x\|_0\leq \|x\|_0, \text{for every $s\in[0,1].$}$$
Then
$$\|A\|=\sup_{x\in C^0} \frac{\|A(x)\|_0}{\|x\|_0}\leq 1.$$
How can I show that $\|A\|\geq 1?$
 A: Your first argument has a mistake; bringing the $s$ out in front doesn't make sense as it's the variable of integration. What we get is
$$\|Af\| =\int_0^1|sf(s)|\, ds \le \int_0^1|f(s)|\,ds = \|f\|,$$
showing $\|A\|\le 1.$ To show $\|A\|\ge 1,$ consider the functions $f_n(s) = s^n, n \in \mathbb N.$
A: No. You are confuse. The correct is $\|Ax\|=\int_0^1|sx(s)|ds$ (which clearly is not $s|x(s)|$. Indeed, the latest has not a precisely meaning). 
Now, as $s|x(s)|\le|x(s)|$ for all $s\in[0,1]$, then $\int_0^1|sx(s)|ds\le\int_0^1|x(s)|ds$. That is, $\|Ax\|\le\|x\|$ for all $x\in C[0,1]$. Then, $\|Ax\|\le 1$ for all $x\in C[0,1]$ such that $\|x\|=1$. Thus, $\|A\|\le 1$. 
Now, set $\epsilon>0$. Take $a\in(0,1)$ such that $1-a/3<\epsilon$. Define $f\in C[0,1]$ by $f(s)=(-\frac{1}{a}s+1)1_{[0,a]}(s)$. Thus:
$\begin{eqnarray*}
\|f\|&=&\int_0^1(-\frac{1}{a}s+1)1_{[0,a]}(s)ds\\
&=&\int_0^a(-\frac{1}{a}s+1)ds\\
&=&\frac{a}{2} 
\end{eqnarray*}$
and
$\begin{eqnarray*}
\|Af\|&=&\int_0^1s(-\frac{1}{a}s+1)1_{[0,a]}(s)ds\\
&=&\int_0^as(-\frac{1}{a}s+1)ds\\
&=&-\frac{1}{a}\int_0^as^2ds+\int_0^asds\\
&=&-\frac{1}{a}\frac{a^3}{3}+\frac{a^2}{2}\\
&=&\frac{a^2}{2}-\frac{a^2}{3}\\
&=&\frac{a^2}{6}
\end{eqnarray*}$
Thus, $\|Af\|/\|f\|=a/3>1-\epsilon$. 
Conclusion: for all $\epsilon>0$ we can exhibit $f\in C[0,1]$ such that $1-\epsilon<\|Af\|/\|f\|<1$. Thus, $1=\sup\{\|Af\|/\|f\|\,\,:\,\,\|f\|\neq 0\}$. This proves $\|A\|=1$.
