[proofplan] Let $W_\Delta := \langle s_\alpha : \alpha \in \Delta\rangle$ be the subgroup generated by the simple reflections. Since $W$ is generated by all root reflections, it is enough to prove that $s_\beta \in W_\Delta$ for every root $\beta \in \Phi$. We do this by reducing a positive root to a simple root: if $\beta$ is positive and not simple, then some simple reflection sends $\beta$ to a positive root of strictly smaller height. Iterating this descent gives an element $w \in W_\Delta$ with $w\beta \in \Delta$, and conjugating the corresponding simple reflection recovers $s_\beta$. [/proofplan]

[step:Reduce the theorem to positive root reflections] Define \begin{align*} W_\Delta := \langle s_\alpha : \alpha \in \Delta\rangle \subseteq W. \end{align*} Since $W$ is generated by the reflections $s_\beta$ with $\beta \in \Phi$, it suffices to prove that $s_\beta \in W_\Delta$ for every $\beta \in \Phi$. Let $\Phi^+$ denote the positive roots determined by the base $\Delta$. If $\beta \in -\Phi^+$, then $-\beta \in \Phi^+$ and \begin{align*} (-\beta)^\vee = -\beta^\vee. \end{align*} Therefore, for every $x \in E$, \begin{align*} s_{-\beta}(x) &= x - \langle x,(-\beta)^\vee\rangle(-\beta) \\ &= x - \langle x,-\beta^\vee\rangle(-\beta) \\ &= x - \langle x,\beta^\vee\rangle\beta \\ &= s_\beta(x). \end{align*} Thus $s_{-\beta} = s_\beta$, and it is enough to prove $s_\beta \in W_\Delta$ for every $\beta \in \Phi^+$. [/step]

[step:Find a simple reflection that lowers the height of a non-simple positive root]Let $\beta \in \Phi^+$ be a positive root which is not simple. Write its expansion in the simple root basis as \begin{align*} \beta = \sum_{\alpha \in \Delta} m_\alpha \alpha, \end{align*} where each $m_\alpha \in \mathbb{Z}_{\geq 0}$ and at least two coefficients are involved, or one coefficient is greater than $1$. Define the height of $\beta$ by \begin{align*} \operatorname{ht}(\beta) := \sum_{\alpha \in \Delta} m_\alpha. \end{align*} We claim that there exists $\alpha_0 \in \Delta$ such that \begin{align*} \langle \beta,\alpha_0^\vee\rangle > 0. \end{align*} Indeed, if $\langle \beta,\alpha^\vee\rangle \leq 0$ for every $\alpha \in \Delta$, then using the expansion of $\beta$ gives \begin{align*} 2(\beta,\beta)_E &= 2\left(\beta,\sum_{\alpha \in \Delta} m_\alpha \alpha\right)_E \\ &= \sum_{\alpha \in \Delta} m_\alpha\,2(\beta,\alpha)_E \\ &= \sum_{\alpha \in \Delta} m_\alpha\,(\alpha,\alpha)_E\langle \beta,\alpha^\vee\rangle \\ &\leq 0. \end{align*} This contradicts $(\beta,\beta)_E > 0$. Hence such an $\alpha_0$ exists. Set \begin{align*} c := \langle \beta,\alpha_0^\vee\rangle. \end{align*} The root-system integrality axiom gives $c \in \mathbb{Z}$, so $c \in \mathbb{N}$. Since $\beta \neq \alpha_0$, the standard positivity property of a base says that $s_{\alpha_0}$ sends the positive root $\beta \neq \alpha_0$ to another positive root. Thus \begin{align*} s_{\alpha_0}\beta = \beta - c\alpha_0 \in \Phi^+. \end{align*} Moreover, \begin{align*} \operatorname{ht}(s_{\alpha_0}\beta) = \operatorname{ht}(\beta - c\alpha_0) = \operatorname{ht}(\beta) - c < \operatorname{ht}(\beta). \end{align*}[/step]

[guided]The descent step needs two facts: first, some simple root must pair positively with $\beta$; second, reflecting across that simple root lowers the height without leaving the positive chamber of roots. Write the positive root $\beta$ in the simple root basis: \begin{align*} \beta = \sum_{\alpha \in \Delta} m_\alpha \alpha, \end{align*} with $m_\alpha \in \mathbb{Z}_{\geq 0}$. Its height is \begin{align*} \operatorname{ht}(\beta) := \sum_{\alpha \in \Delta} m_\alpha. \end{align*} Suppose, toward a contradiction, that every simple coroot pairs non-positively with $\beta$: \begin{align*} \langle \beta,\alpha^\vee\rangle \leq 0 \qquad\text{for every }\alpha \in \Delta. \end{align*} Then the inner product of $\beta$ with itself can be computed from the simple-root expansion: \begin{align*} 2(\beta,\beta)_E &= 2\left(\beta,\sum_{\alpha \in \Delta} m_\alpha \alpha\right)_E \\ &= \sum_{\alpha \in \Delta} m_\alpha\,2(\beta,\alpha)_E \\ &= \sum_{\alpha \in \Delta} m_\alpha\,(\alpha,\alpha)_E\langle \beta,\alpha^\vee\rangle. \end{align*} Each factor $m_\alpha$ is non-negative, each $(\alpha,\alpha)_E$ is positive, and each $\langle \beta,\alpha^\vee\rangle$ is non-positive by assumption. Hence the final sum is at most $0$, so $2(\beta,\beta)_E \leq 0$. This is impossible because $\beta$ is a non-zero vector in a Euclidean space. Therefore there exists $\alpha_0 \in \Delta$ with \begin{align*} \langle \beta,\alpha_0^\vee\rangle > 0. \end{align*} Define \begin{align*} c := \langle \beta,\alpha_0^\vee\rangle. \end{align*} The root-system integrality axiom gives $c \in \mathbb{Z}$, and the positivity just proved gives $c \in \mathbb{N}$. The reflection formula gives \begin{align*} s_{\alpha_0}\beta = \beta - c\alpha_0. \end{align*} Because $\beta$ is positive and $\beta \neq \alpha_0$, the defining positivity property of a base implies that the simple reflection $s_{\alpha_0}$ sends $\beta$ to a positive root. Thus \begin{align*} s_{\alpha_0}\beta \in \Phi^+. \end{align*} Finally, subtracting $c\alpha_0$ reduces the sum of the coefficients in the simple-root expansion by exactly $c$: \begin{align*} \operatorname{ht}(s_{\alpha_0}\beta) = \operatorname{ht}(\beta - c\alpha_0) = \operatorname{ht}(\beta) - c < \operatorname{ht}(\beta). \end{align*} So a non-simple positive root can always be moved by a simple reflection to a positive root of strictly smaller height.[/guided]

[step:Iterate the height reduction until the root becomes simple] Let $\beta \in \Phi^+$. If $\beta \in \Delta$, no reduction is needed. If $\beta \notin \Delta$, the previous step gives $\alpha_1 \in \Delta$ such that $s_{\alpha_1}\beta$ is positive and has smaller height. Repeating this procedure produces positive roots of strictly decreasing positive integer height. Since there is no infinite strictly decreasing sequence in $\mathbb{N}$, the process terminates. Thus there exist simple roots $\alpha_1,\dots,\alpha_k \in \Delta$ and a simple root $\alpha \in \Delta$ such that, with \begin{align*} w := s_{\alpha_k}\cdots s_{\alpha_1} \in W_\Delta, \end{align*} we have \begin{align*} w\beta = \alpha. \end{align*} [/step]

[step:Conjugate the simple reflection back to the original root reflection] Let $\beta \in \Phi^+$. Choose $w \in W_\Delta$ and $\alpha \in \Delta$ such that $w\beta = \alpha$, as constructed in the previous step. For any root $\gamma \in \Phi$ and any element $u \in W$, the reflection in the root $u\gamma$ satisfies \begin{align*} s_{u\gamma} = u s_\gamma u^{-1}. \end{align*} Applying this identity with $u = w$ and $\gamma = \beta$ gives \begin{align*} s_\alpha = s_{w\beta} = w s_\beta w^{-1}. \end{align*} Therefore \begin{align*} s_\beta = w^{-1}s_\alpha w. \end{align*} Since $w \in W_\Delta$, $w^{-1} \in W_\Delta$, and $s_\alpha \in W_\Delta$, the subgroup property gives $s_\beta \in W_\Delta$. By the first step, this also handles negative roots because $s_{-\beta} = s_\beta$. Hence every root reflection $s_\beta$ with $\beta \in \Phi$ lies in $W_\Delta$. Since $W$ is generated by all root reflections, we conclude \begin{align*} W = W_\Delta = \langle s_\alpha : \alpha \in \Delta\rangle. \end{align*} [/step]