Ok, you said you knew how to prove it. Others may not. And you may like this better than what you have (it's the second thing I always think of when this comes up, and I like it a lot better than the first thing I think of...)
Say $u$ is a distribution and $u'=0$. By definition $u(\phi')=0$ for any test function $\phi$. Hence $u(\phi)=0$ for any test function $\phi$ with $\int\phi=0$.
Fix $\psi_0$ with $\int\psi_0=1$, and let $c=u(\psi_0)$. Now for an arbitrary test function $\phi$, let $\alpha=\int\phi$. Then $u(\alpha\psi_0-\phi)=0$, which says $$u(\phi)=c\int\phi.$$Which is exactly what "$u=c$" means.
Detail: We used the following fact above: Given a test function $\phi$ on $\Bbb R$, there exists a test function $\psi$ with $\phi=\psi'$ if and only if $\int\phi=0$. In case this is not clear: First, if $\phi=\psi'$ then $\int\phi=\int\psi'=0$ because $\psi$ has compact support. Suppose on the other hand that $\int\phi=0$, and define $\psi(x)=\int_{-\infty}^x\phi$. Then $\psi'=\phi$ and hence $\psi$ is infinitely differentiable, while the fact that $\int\phi=0$ shows that $\psi$ has compact support.