[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*}
[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]
[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]
