Let $D$ denote the space of all compactly supported infinitely differentiable functions on $\mathbb{R}$, and suppose $D_m$ is the subspace of all infinitely differentiable functions on $\mathbb{R}$ supported on $[-m,m]$ for each $m \in \mathbb{N}$. We can make each $D_m$ a countably normed space by considering the following sequence of norms on $D_m$: $||\phi||^{n}_{m} = \max_{i=0...n} \sup_{t \in [-m,m]} |\phi^{(i)}(t)|$ where $\phi^{(i)}$ denotes the $i$-th derivative, with $\phi^{(0)} = \phi$, and the sequence is $\{|| . ||_{m}^n \}_{n \in \mathbb{N}}$. Now each space $D_m$ with respect to the topology generated by the countable sequence of norms is metrizable, namely via the metric $\rho_{m}(.,.) = \sum_{n=0}^{\infty} 2^{-n} \frac{|| . - . ||^{n}_{m}}{1 + || . - . ||^{n}_{m}}$, and it is easy to see that each of the inclusion maps $i_m : D_m \rightarrow D_{m+1}$ are continuous when each $D_j$ is equipped with the topology generated by this metric. Endow $D$ with the final topology with respect to the inclusion maps $j_m : D_m \rightarrow D$ for each $m$.
I would like to show that: A sequence $\{\phi_n \} \subset D$ converges to $\phi \in D$ if and only if:
(i) There is some $m \in \mathbb{N}$ for whih all $\phi_n,\phi \in D_m$
(ii) $\phi_n^{(k)}$ converges uniformly to $\phi^{(k)}$ for all $k$
I am a bit lost as to how to proceed, I thought about trying to prove that $D$ was metrizable with a metric related to each of the metrics of $D_m$ but this does not seem to be necessarily true, intuitively it seems true but I can't formalize an argument. I know the result itself would be true if I could for instance show each $D_m$ is open in $D$, but this also seems false.
Edit: I think one side of the iff (the if part) can be shown as follows: Since $\phi_n^{(k)} \rightarrow \phi^{(k)}$ uniformly for all $k$, $\phi_n \rightarrow \phi$ in $D_m$, now let $U$ be an open neighbourhood of $\phi$ in $D$, then for $n$ sufficiently large, $\phi_n$ is contained in $U \cap D_m \subset U$, hence $\phi_n \rightarrow \phi$ in $D$. I am still struggling to show the only if side.
It looks like one could metrize $D$ by $\rho(f,g)=\inf_{m\in\mathbb N}\rho_m(f,g)$ (setting $\rho_m(f,g)=\infty$ if $f\notin D_m$ or $g\notin D_m$). This distance means that $\phi_n\to\phi$ in $(D,\rho)$ if and only if $\phi_n\to\phi$ in some $D_m$. As the inclusion maps $j_m\colon(D_m,\rho_m)\to (D,\rho)$ are still continuous, the final topology on $D$ contains the one induced by $\rho$. This should be enough to prove the if part.