How one proves that a certain Fréchet space is a Montel space (such spaces are, unsurprisingly, called Fréchet–Montel spaces, or FM-spaces for short) depends of course on the space in question. Most of the interesting FM-spaces one comes across (at least if one doesn't specialise in functional analysis) are spaces of smooth functions, and for those it's a common theme that the proof employs the Ascoli–Arzelà theorem (more precisely, a generalisation thereof) in a critical way. For spaces of holomorphic functions one often uses Montel's theorem instead, but usually the proof of Montel's theorem uses Ascoli–Arzelà, so these proofs are not very much different in the sense of depending on Ascoli–Arzelà.
Naturally, a different strategy is used to prove that every closed subspace $F$ of an FM-space $E$ is an FM-space (if $K$ is closed and bounded in $F$, then it is closed and bounded in $E$, hence it is compact) or that a countable product of FM-spaces is an FM-space (if $B$ is bounded in $\prod E_k$ then there are closed and bounded sets $B_k \subset E_k$ with $B \subset \prod B_k$, the latter is compact per Tychonov, so $B$ is relatively compact). However, a quotient (by a closed subspace) of an FM-space need not be reflexive, in that case it cannot have the Montel property (aka the Heine–Borel property). But the typical FM-spaces encountered in analysis, like $\mathcal{S}(\mathbb{R}^n)$, $\mathcal{E}(U)$, and $\mathscr{O}(U)$, are not only FM-spaces, they are in fact Fréchet–Schwartz spaces (FS-spaces, this implies the Montel property) and the quotient of a Schwartz space by a closed subspace is again a Schwartz space, so the aforementioned spaces also have nice quotients.
As an example, I'll give a proof that $\mathcal{S}(\mathbb{R}^n)$ has the Montel property. It is convenient to prove that a closed and bounded subset of $\mathcal{S}(\mathbb{R}^n)$ is sequentially compact. Since we're dealing with metrisable spaces sequential compactness and compactness coincide.
I'll use Greek letters for multi-indices and write $D^{\alpha}f$ for
$$\biggl(\frac{\partial}{\partial x_1}\biggr)^{\alpha_1} \dotsb \biggl(\frac{\partial}{\partial x_n}\biggr)^{\alpha_n}f$$
and use the seminorms
$$p_{k,m}(f) = \max_{\lvert\alpha\rvert = k} \sup_{x \in \mathbb{R}^n}\: (1 + \lVert x\rVert^2)^m\bigl\lvert D^{\alpha}f(x)\bigr\rvert$$
to induce the topology of $\mathcal{S}(\mathbb{R}^n)$. It shouldn't be difficult to show that this family is equivalent to whichever family of seminorms are used elsewhere. I use $B_{k,m}$ to denote the closed unit ball of $p_{k,m}$, i.e.
$$B_{k,m} = \{ f \in \mathcal{S}(\mathbb{R}^n) : p_{k,m}(f) \leqslant 1\}\,.$$
Then $B \subset \mathcal{S}(\mathbb{R}^n)$ is bounded if and only if there is a family $(C_{k,m})$ of positive real numbers such that
$$B \subset \bigcap_{k,m} C_{k,m}\cdot B_{k,m}\,,$$
hence we may as well assume that $B$ equals the right hand side.
Let $(f_{\nu})_{\nu \in \mathbb{N}}$ be a sequence in $B$.
Now it is immediate that $\lvert f(x)\rvert \leqslant C_{0,0}$ for all $f \in B$ and all $x \in \mathbb{R}^n$, so $B$ is a uniformly bounded family of functions. Also, since
$$\biggl\lvert \frac{\partial f}{\partial x_i}(x)\biggr\rvert \leqslant C_{1,0}$$
for all $f \in B$, $1 \leqslant i \leqslant n$, and all $x \in \mathbb{R}^n$, by the mean value theorem the family $B$ is equilipschitz, in particular it is a uniformly equicontinuous family. Thus by Ascoli–Arzelà there is a strictly increasing $\sigma_0 \colon \mathbb{N} \to \mathbb{N}$ such that the sequence $\bigl(f_{\sigma_0(\nu)}\bigr)_{\nu \in \mathbb{N}}$ converges compactly to some $g_0 \in C(\mathbb{R}^n)$.
Since all second partial derivatives of members of $B$ are bounded (in absolute value) by $C_{2,0}$ it follows that also the family
$$\{ D^{\alpha} f : f \in B, \lvert\alpha\rvert = 1\}$$
is uniformly equicontinuous (even equilipschitz). And it is uniformly bounded (by $C_{1,0}$). Hence there is a strictly increasing $\sigma_1 \colon \mathbb{N} \to \mathbb{N}$ such that the sequence $\bigl(D^{\alpha}f_{\sigma_0(\sigma_1(\nu))}\bigr)$ converges compactly to some $g_{\alpha} \in C(\mathbb{R}^n)$ for every multi-index $\alpha$ with $\lvert\alpha\rvert = 1$. It follows that $g_0 \in C^1(\mathbb{R}^n)$ and $D^{\alpha}g_0 = g_{\alpha}$ for $\lvert\alpha\rvert = 1$. Set $s_1 = \sigma_0 \circ \sigma_1$.
Having found a strictly increasing $s_r \colon \mathbb{N} \to \mathbb{N}$ such that the sequence $\bigl(D^{\alpha}f_{s_r(\nu)}\bigr)$ converges compactly to some $g_{\alpha} \in C(\mathbb{R}^n)$ for all multi-indices $\alpha$ with $\lvert\alpha\rvert \leqslant r$, we note that $\{ D^{\beta}f : f \in B, \lvert\beta\rvert = r+1\}$ is uniformly bounded by $C_{r+1,0}$, and it is also a uniformly equicontinuous family because $\{ D^{\gamma}f : f \in B, \lvert \gamma\rvert = r+2\}$ is uniformly bounded by $C_{r+2,0}$. Hence (once more by Ascoli–Arzelà) there is a strictly increasing $\sigma_{r+1} \colon \mathbb{N} \to \mathbb{N}$ such that the sequence $\bigl(D^{\alpha}f_{s_{r+1}(\nu)}\bigr)$, where $s_{r+1} = s_r \circ \sigma_{r+1}$, converges compactly to $g_{\alpha} \in C(\mathbb{R}^n)$ for all $\alpha$ with $\lvert\alpha\rvert \leqslant r+1$. It follows that $g_0 \in C^{r+1}(\mathbb{R}^n)$ and $D^{\alpha}g_0 = g_{\alpha}$ for $\lvert\alpha\rvert \leqslant r+1$.
Continue that ad infinitum. Then the diagonal sequence $\bigl(f_{s_{\nu}(\nu)}\bigr)$ is a subsequence of the original sequence such that $\bigl(D^{\alpha}f_{s_{\nu}(\nu)}\bigr)$ converges compactly to some $g_{\alpha} \in C(\mathbb{R}^n)$ for every multi-index $\alpha$. Hence $g_0 \in C^{\infty}(\mathbb{R}^n)$ and $D^{\alpha}g_0 = g_{\alpha}$ for all $\alpha$.
For better readability, define $h_{\nu} = f_{s_{\nu}(\nu)}$.
Next, for arbitrary $k$ and $m$, consider a multi-index $\alpha$ with $\lvert\alpha\rvert = k$ and an arbitrary $x\in \mathbb{R}^n$. Then
$$(1 + \lVert x\rVert)^m\bigl\lvert D^{\alpha}g_0(x)\bigr\rvert = \lim_{\nu \to \infty} (1 + \lVert x\rVert^2)^m\bigl\lvert D^{\alpha}h_{\nu}(x)\bigr\rvert \leqslant C_{k,m}\,.$$
Thus $g_0 \in B \subset \mathcal{S}(\mathbb{R}^n)$.
Finally, we show that $(h_{\nu})$ converges to $g_0$ in $\mathcal{S}(\mathbb{R}^n)$, which completes the proof. Fix arbitrary $k,m$ and choose $\varepsilon > 0$. Pick $R >0$ such that
$$\frac{C_{k,m+1}}{1 + R^2} < \frac{\varepsilon}{2}\,.$$
For $\lVert x\rVert > R$ we have
$$(1 + \lVert x\rVert^2)^m\bigl\lvert D^{\alpha}(g_0 - h_{\nu})(x)\bigr\rvert \leqslant \frac{p_{k,m+1}(g_0) +p_{k,m+1}(h_{\nu})}{1 + \lVert x\rVert^2} < \varepsilon$$
for every $\nu$. By the compact convergence (this is actually uniform on all of $\mathbb{R}^n$) there is a $\nu_0$ (a priori depending on $\alpha$, but since there are only finitely many $\alpha$ with $\lvert \alpha\rvert = k$ we can pick one that works for all these $\alpha$) such that
$$\bigl\lvert D^{\alpha}(g_0 -h_{\nu})(x)\bigr\rvert \leqslant \frac{\varepsilon}{(1 + R^2)^m}$$
for all $\nu \geqslant \nu_0$ and all $x$ with $\lVert x\rVert \leqslant R$. It follows that $p_{k,m}(g_0 - h_{\nu}) \leqslant \varepsilon$ for all $\nu \geqslant \nu_0$. QED
The proof that $\mathcal{E}(U)$, the space of infinitely often differentiable functions on the open subset $U \subset \mathbb{R}^n$ with the topology of compact convergence of all derivatives, is an FM-space is very similar. The uniform boundedness of all derivatives of order $k+1$ of the $f \in B$ (where $B\subset \mathcal{E}(U)$ is bounded) on a neighbourhood of a compact $K \subset U$ implies that the derivatives of order $k$ are a uniformly equicontinuous family on $K$. Since it's also a uniformly bounded family, from every sequence in $B$ we can extract a subsequence such that the derivatives of order $k$ converge compactly. The diagonal sequence converges in $\mathcal{E}(U)$. For the space $\mathscr{O}(U)$ of holomorphic functions on the open $U \subset \mathbb{C}^n$ one can use Montel's theorem or note that it is a closed subspace of $\mathcal{E}(U)$ and use the proposition mentioned at the top to conclude.
Best Answer
A set in a metric space is compact iff it is complete and totally bounded. Here totally bounded means that for any $\varepsilon>0$, there exist finitely many open balls of radius $\varepsilon$ that cover the set. In $\mathbb R^n$, every bounded set is totally bounded (this would be a good exercise to prove).
In infinite dimensional Banach spaces, this is no longer the case. Take $S = \{e_n : n=1,2,\ldots\}\subset \ell^\infty(\mathbb R)$, where $e_n$ is the $n^{\mathrm{th}}$ standard orthonormal basis vector (when viewed in $\ell^2$). Then clearly $S$ is bounded as $\|x\|_\infty = 1$ for all $x\in S$, but for all $x,y\in S$ we have $d(x,y)=1$, so for $\varepsilon=1$ there can be no finite collection of $\varepsilon$ balls that cover the set. Hence $S$ is not totally bounded.