Let $\varphi\in C_c^\infty(\mathbb R)$ and $K:=\operatorname{supp}\varphi$. By the mean value theorem, $$\forall-\infty<a<b<\infty:\exists c\in(a,b):\varphi'(c)=\frac{\varphi(b)-\varphi(a)}{b-a}\tag1$$ and hence $$\left|\varphi(a)-\varphi(b)\right|\le\sup_{c\in\mathbb R}\left|\varphi'(c)\right|\left|a-b\right|\;\;\;\text{for all }a,b\in\mathbb R\tag2.$$ So, $$\left|\varphi(a)\right|\le\left|\varphi(0)\right|+\sup_{c\in\mathbb R}\left|\varphi'(c)\right|\sup_{x\in K}\left|x\right|\;\;\;\text{for all }a\in\mathbb R\tag3.$$
However, I've read that $\left|\varphi\right|$ is actually bounded by $\sup_{x\in\mathbb R}\left|\varphi'(x)\right|$. How can we show that?
I am pretty sure that you need to factor in the size of the support somehow.
Else, given some function $\phi$, you can consider $\phi_r:x \longmapsto \phi(rx)$ where $r>0$ is very small. Every $\phi_r$ is smooth compactly supported and has max-norm the max-norm of $\phi$; the max-norm of $\phi_r’$ is $r$ times the max-norm of $\phi’$ so can be arbitrarily small.
Now, if $\phi$ has support in $[-1,1]$, and if $|a| \leq 1$, there exists $r$ such that $|a-r| \leq 1$ and $\phi(r)=0$. So we know that $\phi(a)-\phi(r)$ can be written as $(a-r)\phi’(c)$ for some $c$, so that $|\phi(a)| \leq 1*|\phi’(c)| \leq \|\phi’\|_{\infty}$.