In Shubin's 'Differential operators and spectral theory' on page 16 he states that linear differential operators are properly supported as pseudo differential operators since the support of their kernel is precisely the diagonal.
I understand why linear differential operators are pseudo differential operators and why if the support of the kernel is the diagonal this would imply that they are properly supported.
However I do not see why the support of the kernel should be the diagonal.
You can start by looking at the Schwarz kernel of the identity operator (it is a Dirac delta distribution). We can in particular write:
\begin{equation} u(x)=(\delta \ast u )(x)=\int \delta(x-y)u(y)dy \end{equation}
You can formally differentiate at both sides of that equality to obtain
\begin{equation} \partial_ju(x)=\int \partial_j\delta(x-y)u(y)dy \end{equation}
which is also suggested by the formula:
\begin{equation} \delta'\ast u=\delta \ast u' = u' \end{equation} (for compactly supported smooth functions, but let us omit technicalities).
Therefore, the Schwarz kernel of the operator derivative in the $j$-coordinate is $\partial_j\delta(x-y)$.
Use linearity to generalise to differential operators, so that operators constructed by finite sums $\sum_\alpha a_{\alpha}(x)D^{\alpha}$, where each $\alpha$ is a multi-index of order smaller than the order of the differential operator the sum represents, has kernel $\sum_\alpha a_{\alpha}D^{\alpha}\delta(x-y)$.
The delta distribution is only non-vanishing at $0$, meaning we require $x=y$, and you have your result.