The hyperplanes are in bijection with Coxeter reflections

Each conjugate $ws{w}^{-1}$ is an involution with fixed hyperplane $w\mathrm{Fix}(s)$, hence is a reflection. Conversely, let $r\in W$ be a reflection with hyperplane $H$. The family $\{H\cap K:K\in \mathcal{A}(W),K\mathrm{\ne }H\}$ is finite because $W$ is finite. Each member is a proper subspace of $H$, since distinct hyperplanes have equal dimension and cannot contain one another. Apply [F3] in the finite-dimensional real vector space $H$ to choose $x\in H$ outside every other arrangement hyperplane. The existence of the hyperplane $H$ implies $n:=\dim E\ge 1$.

Use the coordinate isomorphism $c:E\to {\mathbb{R}}^n$ and metric of [F7]. For each of the finitely many $K\in \mathcal{A}(W)$ choose $u_K\mathrm{\notin }K$. Codimension one gives a unique decomposition $z=k+t{u}_K$ with $k\in K$; define ${\beta }_K(z)=t$. This is a nonzero real linear functional with kernel $K$. Applying [F8] to ${\beta }_K\circ c^{-1}$ gives a constant $C_K>0$ such that $\mathrm{\mid }{\beta }_K(z)-{\beta }_K(z_0)\mathrm{\mid }\le C_K{d}_E(z,z_0)$. Thus at a point where ${\beta }_K$ is nonzero, a ball of radius $\mathrm{\mid }{\beta }_K(z)\mathrm{\mid }\mathrm{/}(2C_K)$ preserves its strict sign. The positive and negative sign sets are open, so $K$ and each closed sign half-space are closed. The same inequality, with radius $\epsilon \mathrm{/}{C}_K$, proves continuity of ${\beta }_K$. Only finitely many choices were made.

A convex subset $Z\subseteq E$ has convex image $c(Z)$, connected in the Euclidean topology by [F9], since $n\ge 1$. The coordinate metric in [F7] transports relative open sets in both directions; a separation of $Z$ would therefore give a separation of $c(Z)$, contrary to [F10]. Thus convex subsets of $E$ are connected, and its metric balls are convex by [F9]. Choose $\rho >0$ smaller than every $\mathrm{\mid }{\beta }_K(x)\mathrm{\mid }\mathrm{/}(2C_K)$ for $K\mathrm{\ne }H$, taking $\rho =1$ if that family is empty. The ball $B=B_E(x,\rho )$ then misses every $K\mathrm{\ne }H$ by step 2.1. Its two strict ${\beta }_H$ half-balls are nonempty: $x\pm t{u}_H$ lie in them for sufficiently small $t>0$. Each half-ball is convex, hence connected. Fix one, $B_{+}$, and $y\in B_{+}$. The component $D$ of the arrangement complement containing $y$ contains $B_{+}$ by [F10]. For $z\in B\cap H$ and $0<t\le 1$, the point $(1-t)z+ty$ belongs to $B_{+}$ by convexity and the strict sign of ${\beta }_H(y)$; its distance to $z$ is $t{d}_E(y,z)$. Hence every neighbourhood of $z$ meets $B_{+}$, giving $B\cap H\subseteq \bar{D}$.

Orient each ${\beta }_K$ from step 2.1 by a sign to obtain ${\alpha }_K(y)>0$. Its sign is positive throughout $D$: otherwise the two relatively open strict sign sets would separate $D$, since $D$ misses $K$, contradicting [F10]. Thus $D\subseteq P:=\{z\in E:{\alpha }_K(z)>0\text{ for every }K\in \mathcal{A}(W)\}$. Conversely, $P$ is convex, hence connected by step 3.1, lies in the arrangement complement and contains $y$. The component property in [F10] gives $P\subseteq D$, so $D=P$. It is open as a finite intersection of strict sign sets. Put $Q=\{z\in E:{\alpha }_K(z)\ge 0\text{ for every }K\in \mathcal{A}(W)\}$. The closed sign half-spaces from step 2.1 show that $Q$ is closed and contains $\bar{D}$. For $z\in Q$ and $0<t\le 1$, every ${\alpha }_K((1-t)z+ty)$ is positive, so $(1-t)z+ty\in D$. Its distance to $z$ is $t{d}_E(y,z)$, proving $z\in \bar{D}$. Therefore $\bar{D}=Q$, a closed convex cone, full-dimensional because it contains the nonempty open set $D$.

Put $F=Q\cap H$. A face of a convex set $Q$ means a convex subset $F$ such that, whenever $u,v\in Q$, $0<t<1$ and $(1-t)u+tv\in F$, both $u$ and $v$ belong to $F$. Here $F$ is convex, and ${\alpha }_H\ge 0$ on $Q$. If $(1-t)u+tv\in F$, then $0=(1-t){\alpha }_H(u)+t{\alpha }_H(v)$ forces both nonnegative terms to vanish; hence $u,v\in H$ and therefore $u,v\in F$. This proves that $F$ is a face. Moreover, $F\subseteq H$ and step 3.1 gives $B\cap H\subseteq F$. This relative open ball has affine hull $H$ (its point differences span $H$), so $\dim \mathrm{aff}(F)=\dim H=\dim E-1$. Thus $F$ is a codimension-one face of the full-dimensional $\bar{D}$, and $H$ is a wall hyperplane in the convention of [F1].

Choose $w$ with $D=wC$ by [F1]. Applying [F8] in coordinates to $w$ and $w^{-1}$ makes both continuous. Thus $w^{-1}\bar{D}=\bar{C}$. Invertible linear maps preserve convex combinations and affine dimension, so the face in step 5.1 pulls back to a codimension-one face of $\bar{C}$. Hence $w^{-1}H$ is a wall hyperplane of $C$, and its reflection is some $s\in S$. By uniqueness in [F2], $r=ws{w}^{-1}$. This proves equality of the two reflection sets. Surjectivity onto $\mathcal{A}(W)$ holds by its definition and injectivity by [F2], so this map is a bijection by [F4]. Since $T\subseteq W$ is finite, [F5] gives $\mathrm{\mid }\mathcal{A}(W)\mathrm{\mid }=\mathrm{\mid }T\mathrm{\mid }$.