Any such topology will be fairly unpleasant. For instance, the topology of $X$ cannot be induced by any translation-invariant metric $d$.
Lemma. Let $Y_1, Y_2$ be two topological vector spaces whose topologies are induced by translation-invariant metrics $d_1, d_2$, and let $T : Y_1 \to Y_2$ be a continuous linear map. Then $T$ is uniformly continuous.
Proof. Since $T$ is continuous at 0, for any $\epsilon > 0$ there exists $\delta > 0$ such that if $d_1(x, 0) < \delta$ then $d_2(Tx, 0) < \epsilon$. Now if $d_1(x,y) < \delta$, then $d_1(x-y, 0) = d_1(x,y) < \delta$ and we have $d_2(Tx, Ty) = d_2(Tx-Ty, 0) = d_2(T(x-y), 0) < \epsilon$.
Now recall that $C(\mathbb{R}^n; \mathbb{R}^d)$ is a Fréchet space, so its usual topology is induced by a complete translation-invariant metric $d_0$. By assumption, the identity map $id$ from $(C(\mathbb{R}^n; \mathbb{R}^m), d_0)$ to $(C(\mathbb{R}^n; \mathbb{R}^m), d)$ is a homeomorphism, and so by our lemma, $id$ and $id^{-1}$ are uniformly continuous. In particular, $C(\mathbb{R}^n; \mathbb{R}^m)$ is complete with respect to $d$, and therefore closed in $X$.