If $\dim {\mathfrak g}_{-\alpha}>1$, then the map $[-,X_\alpha]: {\mathfrak g}_{-\alpha}\to {\mathfrak h}_\alpha$ can't be injective since ${\mathfrak h}_\alpha$ is $1$-dimensional as you already know. Explicitly, if $Z\in{\mathfrak g}_{-\alpha}$ then $Z^{\prime} := Z - [Z,X_{\alpha}]X_{-\alpha}$ satisfies $[Z,X_\alpha]=0$.
Alternatively, the proof I know and like is the following: Consider $${\mathfrak g}(\alpha) := {\mathbb k}\cdot X_{-\alpha}\oplus {\mathfrak h}_{\alpha}\oplus\bigoplus\limits_{n\geq 1} {\mathfrak g}_{n\alpha}.$$ It is naturally a module over the Lie subalgebra ${\mathbb k}\cdot\{X_{-\alpha},H_\alpha,X_\alpha\}\cong{\mathfrak sl}_2({\mathbb k})$ of ${\mathfrak g}$. Hence, since $H_\alpha = [X_\alpha,X_{-\alpha}]$, the trace of the adjoint action $[H_\alpha,-]$ on ${\mathfrak g}(\alpha)$ must be $0$, but on the other hand equals $$-2 + 0 + 2\cdot \dim {\mathfrak g}_\alpha + 4\cdot\dim {\mathfrak g}_{2\alpha} + ...$$
It follows that $\dim{\mathfrak g}_\alpha=1$ and $n\alpha\notin\Phi$ for all $n>1$.
In the infinite-dimensional setting of Kac-Moody algebras things break down very early, since it is not true in general that $[{\mathfrak h}_\alpha,{\mathfrak g}_\alpha] = [[{\mathfrak g}_\alpha,{\mathfrak g}_{-\alpha}],{\mathfrak g}_\alpha]\neq 0$, preventing you from constructing an ${\mathfrak sl}_2$-triple from any root (it only works for the real roots).
Took some work and some scouring through the literature, but we got there in the end. Both questions are answered positively — though my brain still needs some time to digest the answer for question one, and make sure that it's really really true, so, eh, approach with care.
First question:
Whenever I say parabolic here, I mean not the Knapp definition, but the one in terms of complexification, see the OP.
From Lemma in Section 3.2 from Wolf, Koranyi, we can extract the following (heavily paraphrased, but hopefully equivalent):
Let $\mathfrak{g}$ be a real semisimple Lie algebra and $\mathfrak{q}$ a parabolic subalgebra. Then there is some Cartan decomposition
$$\mathfrak{g} = \mathfrak{t} \oplus \mathfrak{p} $$
some maximally noncompact ("maximally split") Cartan subalgebra $\mathfrak{h} = \mathfrak{t} \oplus \mathfrak{a}$ of $\mathfrak{g}$, where
$$\mathfrak{t} \subset \mathfrak{k} \text{ (the "totally nonsplit" part)}, \quad \mathfrak{a} \subset \mathfrak{p} \text{ (the "totally split" part) },$$
a subspace $\mathfrak{a}' \subset \mathfrak{a}$ and a choice of positive roots $P$ in the restricted root space decomposition of $(\mathfrak{g}, \mathfrak{a}')$ so that
$$\mathfrak{q} = \mathfrak{g}_0 \oplus \bigoplus_{\alpha \in P} \mathfrak{g}_\alpha, \quad
\mathfrak{g}_\alpha = \{x \in \mathfrak{g} : [a,x] = \alpha(a) \cdot x \quad \forall a \in \mathfrak{a}'\}.
$$
(Careful: In the source, the root space decomposition is carried out in the complexification $\mathfrak{g}_\mathbb{C}$, but we can also carry it out in the real setting, since the ad-action of elements in $\mathfrak{a}' \subset \mathfrak{a}$ is real diagonalizable. This is always necessary for the restricted root space decomposition.)
Very verbose, but in the end, in the above notation, every parabolic subalgebra contains $Z_\mathfrak{k}(\mathfrak{a}) \oplus \mathfrak{a}$ in the $\mathfrak{g}_0$-component and some choice of positive restricted roots of $\mathfrak{a}$ in the $\bigoplus_{\alpha \in P} \mathfrak{g}_\alpha$-component. Hence every parabolic subalgebra contains some subalgebra of the form $\mathfrak{m} \oplus \mathfrak{a} \oplus \mathfrak{n}$. And indeed, the subalgebras $\mathfrak{m} \oplus \mathfrak{a} \oplus \mathfrak{n}$ are parabolic, since their complexification contains a Borel algebra associated to the complexification of the Cartan subalgebra $\mathfrak{a} \oplus \mathfrak{t}$.
Hence, Knapp's minimal parabolic subalgebras are exactly the minimal parabolic subalgebras in the usual sense.
Second question:
In Bourbaki, Chapter VIII, Exercise 3a for §5, we learn:
If $\mathfrak{q}, \mathfrak{p}$ are parabolic subalgebras of a semisimple (real or complex) Lie algebra $\mathfrak{g}$, and $\mathfrak{q} \subset \mathfrak{p}$, then the radical of $\mathfrak{p}$ is contained in the radical of $\mathfrak{q}$.
And in Bourbaki, Chapter VIII, §10, Corollary 2, we learn:
Every subalgebra $\mathfrak{n}$ of a (real or complex) Lie algebra $\mathfrak{g}$, consisting only of nilpotent elements of $\mathfrak{g}$, is contained in the nilradical of a parabolic subalgebra $\mathfrak{q}$.
As a corollary of the two: Given a subalgebra $\mathfrak{n} \subset \mathfrak{g}$, contained in the radical of some parabolic subalgebra $\mathfrak{q}$. But then there is some minimal parabolic $\mathfrak{q}_0 \subset \mathfrak{q}$ with $\mathfrak{n} \subset \text{rad}(\mathfrak{q}) \subset \text{rad}(\mathfrak{q}_0)$.
Wolf, J. A.; Koranyi, A., Generalized Cayley transformation of bounded symmetric domains, Am. J. Math. 87, 899-939 (1965). ZBL0137.27403.
Bourbaki, Nicolas, Elements of mathematics. Lie groups and Lie algebras. Chapters 7 and 8, Berlin: Springer (ISBN 3-540-33939-6). 271 p. (2006). ZBL1181.17001.
Best Answer
It is true:
Let $\mathcal{K}$ be the Killing form of $\mathfrak{g}$, using invariance of $\mathcal{K}$, one shows that $[\mathfrak{g},e]\subset{\mathfrak{g}^e}^\perp$ (orthogonality is with respect to $\mathcal{K}$). By non degeneracy of $\mathcal{K}$, the two spaces have same dimension hence they are equal. Again using invariance of $\mathcal{K}$ and knowing that $\mathfrak{g}^e$ is abelian because $e$ is regular, we conclude that $[\mathfrak{g}^e,\mathfrak{g}]\subset{\mathfrak{g}^e}^\perp=[\mathfrak{g},e]$.