The root system of a simple Lie algebra is irreducible

Form the graph whose vertices are roots and whose edges join nonorthogonal roots. Suppose it has at least two connected components, and let ${\mathrm{\Phi }}_i$ be the roots in one component. If $\alpha \in {\mathrm{\Phi }}_i$, $\beta \mathrm{\notin }{\mathrm{\Phi }}_i$, then $(\alpha ,\beta )=0$. Moreover $\alpha +\beta$ cannot be a root: if it were, $(\alpha +\beta ,\alpha )=(\alpha ,\alpha )\mathrm{\ne }0$ and $(\alpha +\beta ,\beta )=(\beta ,\beta )\mathrm{\ne }0$, joining both components to the vertex $\alpha +\beta$. Likewise $\beta =-\alpha$ is impossible across components because then $(\alpha ,\beta )=-(\alpha ,\alpha )\mathrm{\ne }0$.

Let ${\mathfrak{h}}_i$ be the span of the coroots of roots in ${\mathrm{\Phi }}_i$ and put ${\mathfrak{g}}_i={\mathfrak{h}}_i\oplus {⨁}_{\alpha \in {\mathrm{\Phi }}_i}{\mathfrak{g}}_{\alpha }$. Partition the simple roots by graph component. A simple reflection from another component fixes a root orthogonal to its simple root; a reflection from the same component either fixes the root or sends it to a root still nonorthogonal to that simple root. Since every root is a Weyl translate of a simple root [F3], the coroots of each component lie in the span of that component's simple coroots. Their basis property gives $\mathfrak{h}={⨁}_i{\mathfrak{h}}_i$. For $\alpha \in {\mathrm{\Phi }}_i$ and $\beta \in {\mathrm{\Phi }}_j$, $i\mathrm{\ne }j$, orthogonality gives $\beta (h_{\alpha })=0$; step 1.1 and [F2] give $[{\mathfrak{g}}_{\alpha },{\mathfrak{g}}_{\beta }]=0$. Thus every ${\mathfrak{g}}_i$ is stable under brackets with $\mathfrak{h}$ and all root spaces, so is an ideal of $\mathfrak{g}$.

Each ${\mathfrak{g}}_i$ is nonzero because it contains a root space. With at least two components it is proper because the root and coroot bases in [F3] split among them. This contradicts simplicity [F1]. Therefore the graph is connected, proving irreducibility.