If the setting is the one of my comment above, I think that $C^k_0(\Omega)$ is not Montel, for finite $k$. Below, for $K\subset \Omega$ a compact set I write $C^k_K(\Omega)$ the set of all $C^k(\Omega)$ functions having support included in $K$. I also fix a countable exhaustion $(K_n)_n$ of $\Omega$.

One property (characterization) of the inductive limit topology on $C^k_0(\Omega)$ is the following : it's the finest LCTVS topology for which the embeddings $C^k_{K_n}(\Omega)\hookrightarrow C^k_0(\Omega)$ are all continuous. But $C^k_{K_n}(\Omega)$ is a normed space of infinite dimension : it cannot be Montel (this arguments fails when $k=\infty$ obviously), for instance the unit ball of $C^k_{K_n}(\Omega)$ is bounded and not compact. Thanks to the previous embedding, this unit ball is also bounded in $C^k_0(\Omega)$ and cannot be compact in it, otherwise this would imply compactness in $C^k_{K_n}(\Omega)$.

In the following I write $C^k_K(\Omega)$ for the space of all $C^k(\Omega)$ functions having support included in $K$, equipped with its "natural" norm $\|\cdot\|_{K,k}$ (uniform norm on $K$ of all derivatives up to the order $k$). I also fix a countable exhaustion $(K_n)_n$ of $\Omega$. I assume that $C^k_0(\Omega)$ is equipped with the inductive limit topology $\tau$ which satisfies in particular :

$(i)$ For all $n$, the embedding $C^k_{K_n}(\Omega)\hookrightarrow C^k_0(\Omega)$ is continuous

$(ii)$ The trace topology of $\tau$ on $C^k_{K_n}(\Omega)$ is precisely the one of $\|\cdot\|_{K_n,k}$.

In this setting, I think that $C^k_0(\Omega)$ is not Montel, for finite $k$.

Indeed : $C^k_{K_n}(\Omega)$ is a normed space of infinite dimension : it cannot be Montel (this argument fails when $k=\infty$ obviously), for instance the unit ball of $C^k_{K_n}(\Omega)$ is bounded and not compact. Because of $(i)$, this unit ball is also bounded in $C^k_0(\Omega)$. But because of $(ii)$, it cannot be compact for the $\tau$ toplogy, otherwise this would imply compactness in $C^k_{K_n}(\Omega)$.