How does one prove that \begin{equation} x T = 0 \Longleftrightarrow T = c \delta_0, \end{equation} with $T \in \mathcal{D}'(\mathbb{R})$?
I see why we have "$\Longleftarrow$ " ($T = c\delta_0 \Leftrightarrow xT = xc\delta_0 \Leftrightarrow \langle xT, \phi \rangle = \langle c\delta_0, x\phi \rangle$ for all $\phi \in \mathcal{D}(\mathbb{R})$ and $\langle \delta_0, x\phi \rangle = 0$ therefore $xT = 0$).
There may be some details to work out, but here's a rough sketch. Let us apply Fourier transform techniques.
\begin{align} 0 &= \langle xT, \varphi\rangle \\ &= \langle T,x\varphi\rangle \\ &= \langle \mathcal{F}T, \mathcal{F}(x\varphi)\rangle \\ &= \langle \mathcal{F}T, (\mathcal{F}\varphi)'\rangle \\ &= -\langle (\mathcal{F}T)',\mathcal{F}\varphi\rangle \end{align}
Thus the distribution $\mathcal{F}T$ has derivative zero. This implies that it is constant and the inverse Fourier transform of a constant distribution is a multiple of the Dirac delta at $0$.
Here are some details for showing that if the derivative of a distribution is zero, it is constant.