Convergence of test-functions is not induced by any metric.

It is a consequence of the Baire category theorem. Essentially, $\mathcal{D}$ is of first category in itself and Cauchy sequences converge in $\mathcal{D}$, and this prevents metrizability. You can find a complete discussion in paragraph 6.9 of the book Functional analysis by Walter Rudin.

As it is quite easy to prove it directly, I will add another answer. Let $(\varphi_n)_n$ be a sequence in $\mathcal{D}$ with $\varphi_n(x)=1$ for $|x|\leq n$ and $\varphi_n(x)=0$ for $|x|>n+1$. Assume $d$ to be a metric on $\mathcal{D}$ compatible with the topology. Let $B_n$ be the ball around $0$ with radius $1/n$ in this metric. As each $B_n$ is absorbing, there is some $c_n>0$ with $c_n\varphi_n \in B_n$ for each $n$. Hence you have $$ c_n\varphi_n \longrightarrow 0 $$ in the metric $d$. By the above definition of the topology, there is a compact set $K\subset\mathbb{R}$ with $\operatorname{supp}c_n\varphi_n \subset K$ for all $n$, which is a contradiction.