Given an isomorphism of two irreducible root systems $\Phi$ and $\Phi$' we need to show that the corresponding simple Lie algebras $L$ and $L'$ are isomorphic. For that we take the subalgebra $D$ of $ L \oplus L'$ generated by $\bar{x_{\alpha}}=(x_{\alpha}, x_{\alpha}')$, $\bar{y_{\alpha}}=(y_{\alpha}, y_{\alpha}')$ and $\bar{h_{\alpha}}=(h_{\alpha}, h_{\alpha}')$ and we show that the projections from $D$ to $L$ and $L'$ are isomorphisms.
The 1st step for this is to show that $D$ is a proper subalgebra. For that we take M the subspace generated by all $ ad \bar{y_{\alpha_1}}.ad \bar{y_{\alpha_2}} \cdots ad \bar{y_{\alpha_m}} (\bar{x})$ where $\bar{x}=(x,x')$, $x$ and $x'$ are elements of the root spaces corresponding to the highest roots $\beta$ and $\beta$' respectively. Now $ ad \bar{y_{\alpha_1}}.ad \bar{y_{\alpha_2}} \cdots ad \bar{y_{\alpha_m}} (\bar{x})$ where $\bar{x}=(x,x')$ belongs to $L_{\beta-\sum \alpha_i} \oplus L'_{\beta-\sum \alpha_i}$. How does it follow from this that $M \cap L_{\beta} \oplus L'_{\beta'}$ is one dimensional ? I guess $L_{\beta} \oplus L'_{\beta'}$ is two dimensional. Again why its immediate that $M$ is a proper subspace ? Humphreys: Lie algebras and Representation theory (page: 75-76).
Note that every element of $M \cap (L_{\beta} \oplus L_{\beta'})$ is a linear combination of elements of the form $ad \bar{y_{\alpha_1}}.ad \bar{y_{\alpha_2}} \cdots ad \bar{y_{\alpha_m}} (\bar{x})$. But to be contained in $L_{\beta} \oplus L_{\beta'}$ every summand in this linear combination has to be contained in $L_{\beta} \oplus L_{\beta'}$ (you have a direct sum decomposition of $L \oplus L'$). As $\bar{x}$ is the only one of these generators being contained in $L_{\beta} \oplus L_{\beta'}$, every element of $M \cap (L_{\beta} \oplus L_{\beta'})$ is a multiple of $\bar{x}$.