Why is $\mathcal E$ a Montel-space? Let $\Omega \subset R^n$ be open, and let us equip $\mathcal E = C^\infty$ with the locally convex topology induced by the following family of semi-norms:
For $K \subset \Omega$ and $\alpha, \beta \in \mathbb N _ 0 ^n$, let $|f|_{K,\alpha,\beta} = sup_{x \in K} | \partial_x^{\alpha} f(x) |$.
It is easy to see that compact sets are closed and bounded within that topology. But how does the opposite direction work?
Thank you!
 A: This is far too long to be a comment, so it's an answer instead.
I don't think anyone usually gives $C^\infty(\Omega)$ another name. Now I presume you've already shown the space to be barreled, so we just need $X \subset C^\infty(\Omega)$ compact iff $X$ is closed and bounded.
First note that your sufficient family of seminorms is too large. Because $\Omega$ is $\sigma$-compact, we can thin out the family by taking a countable family of compact sets $K_j$ that exhaust $\Omega$, i.e. $K_j \subsetneq K_{j+1}$ and $\bigcup_j K_j = \Omega$, so 
$$ \Vert f \Vert_{i,j} = \max_{|\alpha|\leq i} \; \sup_{x \in K_j} |\partial^\alpha f(x)|$$
is a countable family of seminorms that generate the same topology. From these we can define a metric on $C^\infty(\Omega)$ as follows
$$ d(f,g) = \sum_{i=0}^\infty\; \sup_{j \in \mathbb{N}} \frac{\Vert f-g \Vert_{i,j}}{1 + \Vert f-g \Vert_{i,j}} 2^{-i}.$$
It is easy to verify that this metric generates the same topology as the family of seminorms and that the metric is complete and translation-invariant. Thus $C^\infty(\Omega)$ is a Fréchet space and we know that the compact subsets of complete metric spaces are precisely those that are closed and totally bounded. 
So the strategy is clear. First show that $X$ bounded w.r.t. the seminorms implies $X$ bounded w.r.t. the metric. Then show that $X$ bounded w.r.t. the metric implies that $X$ is totally bounded.
