If $x\in[a,b],$ then $|x|\leq \max\{|a|,|b|\}$ The following seems to be quite known: 

If $x\in[a,b],$ then $|x|\leq \max\{|a|,|b|\}$

I do use it while treating problems on series of functions but how does it come about? Any hint please?
 A: If $x\ge0$ then $b\ge0$,  hence $$|x|=x\le b=|b|\le\max(|a|,|b|).$$Similarly if $x<0$ then $a<0$, hence $$|x|=-x\le-a=|a|\le\max(|a|,|b|).$$
A: Since $x \in [a,b]$, it is $a<b$ (or even $a \leq b$ if you want to be more strict for a trivial case).
If $\text{sgn}(a) = \text{sgn}(b)=1$, then $\max\{|a|,|b|\} = |b| \implies |x| \leq \max\{|a|,|b|\}$.
If $\text{sgn}(a) = \text{sgn}(b)=-1$, then $\max\{|a|,|b|\} = |a| \implies |x| \leq \max\{|a|,|b|\}$.
In the case of $\text{sgn}(a) \neq \text{sgn}(b)$, that means that $a$ would be negative. But, in that case, it can be $|a| > |b|$. Then, for $|x|$, it would be : $|x| \leq |a| \implies |x| \leq \max\{|a|,|b|\}$.
So, for all cases, we have that : $|x| \leq \max\{|a|,|b|\}$.
Edit (Alternativelly) : The comment of Jakobian is really nice to consider for a really straightforward proof. What the absolute value $|x|$ means is the distance of $x$ from $0$. But since $x \in [a,b]$, then this distance cannot be larger than the distance of $a$ from zero $(|a|)$ or the distance of $b$ from zero $(|b|)$.
