Prove that every such $f$ is $=0$ everywhere

Suppose that $x\gt c$. First we deal with $c\lt x\lt c+1$. By the Mean Value Theorem, there is an $x_1$ between $c$ and $x$ such that $\frac{f(x)-f(c)}{x-c}=f'(x_1)$. But $f'(x_1)\le f(x_1)\le f(x)$. Thus $f(x)\le (x-c)f(x)$. Since $0\lt x-c\lt 1$, this is only possible if $f(x)=0$.

We conclude that $f(x)=0$ for all $x$ in the interval $(c,c+1)$. By continuity $f(c+1)=0$.

Now we prove in the same way that $f(x)=0$ for all $x$ in the interval $(c+1,c+2]$, and then the interval $(c+2,c+3]$ and so on.

Another Proof :

Let , $f(c)=0$. We want to show $f\equiv 0$ in $\mathbb R=(-\infty,c]\cup[c,\infty)=A\cup B\text{ (say) }$.

$\bullet$ In $A$ , $f'(x)\ge 0$ implies $f$ is monotone increasing in $A$. So , $x\le c\implies f(x)\le f(c)=0$. Also , $f(x)\ge 0$ (given). Hence , $f(x)=0$ in $A$.

$\bullet$ In $B$ , define $F(x)=e^{-x}f(x)$. Then $F'(x)=e^{-x}f'(x)-e^{-x}f(x)\le 0$. So , $F$ is monotone decreasing in $B$. So , $x\ge c\implies F(x)\le F(c)=0\implies f(x)\le0$ , as $e^{-x}>0.$ Also , $f(x)\ge 0$ (given)/ Hence , $f(x)=0$ in $B$.

Consequently , $f\equiv 0$ in $A\cup B=\mathbb R$.