As I have said in comments and the answer to the linked question: If we do not know at this point that the "roots" (defined as non-zero weights of a Cartan subalgebra) form a (reduced) root system, then at the very least one needs the result of Lemma 2 below. The proofs below (and I know no other) need a version of Jacobson-Morozov as well as representation theory of $\mathfrak{sl}_2$. (Which makes me think it is better to use this machinery a little further to show, once and for all, that the roots form a reduced root system; in fact, Lemma 2 shows the "reduced" part in general.)
Let $\alpha, \beta$ be two proportional roots, i.e. there is $c\in \mathbb C$ such that $\beta = c \alpha$.
Lemma 1: $c \in \{\pm \frac12, \pm 1, \pm 2 \}$.
Proof: You say you know there are $x_\alpha \in L_\alpha, y_{-\alpha} \in L_{-\alpha}, h_\alpha \in H$ such that $\alpha(h_\alpha)=2$ and
$$\mathfrak{s}(\alpha) := \mathbb C x_\alpha \oplus \mathbb C y_{-\alpha} \oplus \mathbb C h_\alpha \simeq \mathfrak{sl}_2(\mathbb C).$$
Note that for any $\beta \in \Phi$,
$$\bigoplus_{i \in \mathbb Z} L_{\beta + i \alpha}$$
(where the summands are understood to be $=0$ in case $\beta + i \alpha \notin \Phi \cup \{0\}$) is a $\mathfrak{sl}_2$-representation via the action of $\mathfrak{s}(\alpha)$ on it: It is the $\mathfrak{s}(\alpha)$-module generated by $L_\beta$. In all such representations, all eigenvalues of $h_\alpha$ are integers; but on $L_\beta$, by definition $h_\alpha$ operates via multiplication with $\beta(h_\alpha)$. So $c = \frac{\beta(h_\alpha)}{\alpha(h_\alpha)} \in \frac12 \mathbb Z$.
Now, $\beta = c \alpha \Leftrightarrow \alpha = c^{-1} \beta$, and $\pm \frac12, \pm 1, \pm 2$ are the only elements of $\frac12 \mathbb Z$ whose reciprocals are also in $\frac12 \mathbb Z$.
Lemma 2: If all root spaces $L_\alpha$ are one-dimensional (which they are here), necessarily $c \in \{\pm 1\}$.
Proof: Assume on the contrary that there is $\alpha \in \Phi$ with $2\alpha \in \Phi$. With the notation of Lemma 1, consider the case $\alpha = \beta$, i.e. the $\mathfrak s(\alpha)$-module $$\bigoplus_{i \in \mathbb Z} L_{i \alpha}.$$
$\mathfrak{sl}_2$-representation theory tells us that the maps $[x_\alpha, \cdot ]: L_{i\alpha} \rightarrow L_{(i+1)\alpha}$ are surjective for $i \ge 0$. But if $L_\alpha$ has dimension $1$, it is spanned by $x_\alpha$, so that map is the zero map already for $i=1$.
Corollary: If $\dim(H)=1$, then $\lvert \Phi \rvert =2$.
Note: I am nitpicky in Lemma 2 because in more general situations, over other fields, there can be root spaces of dimension $\ge 2$ and non-reduced root systems, cf. Examples of Lie algebras of the $BC$ root system type or Where does non-reduced root system come up? and links there. You seem to be saying you know already that the root spaces are one-dimensional. That in itself is something for which, to my knowledge, one already needs this $\mathfrak{sl}_2$-machinery. As for non-circular arguments to show it, cf. How to show this $L$-module is simple? (related to the root space decomposition of semisimle Lie algebras), also https://math.stackexchange.com/a/4355610/96384. The issue recently came up in the question Section 2.4 of Samelson's Notes on Lie Algebras.
