[proofplan] We classify the system by looking at its root lines rather than at individual roots. Choose two adjacent root lines and let $\theta$ be the angle between them. The reflection axioms force all root lines to be obtained by repeatedly reflecting across adjacent root lines, so $\theta$ must be of the form $\pi/m$, where $m$ is the number of root lines. The crystallographic condition restricts the possible values of $4\cos^2\theta$ to $0,1,2,3$, giving exactly the four angles $\pi/2,\pi/3,\pi/4,\pi/6$. Finally, reducedness gives exactly two roots on each root line, and the corresponding Cartan integers identify the four cases as $A_1 \times A_1$, $A_2$, $B_2$, and $G_2$. [/proofplan]

[step:Pass from roots to root lines]For each root $\alpha \in \Phi$, define its root line by \begin{align*} L_\alpha := \mathbb{R}\alpha \subset V. \end{align*} Let \begin{align*} \mathcal{L} := \{L_\alpha : \alpha \in \Phi\} \end{align*} be the finite set of root lines. Since $\Phi$ is reduced, the only roots on $L_\alpha$ are $\alpha$ and $-\alpha$. Hence \begin{align*} |\Phi| = 2|\mathcal{L}|. \end{align*} Because $\Phi$ has rank two, $\mathcal{L}$ contains at least two distinct lines and lies in the real projective circle of lines through the origin in $V$. Choose two adjacent lines $L_\alpha,L_\beta \in \mathcal{L}$, meaning that the smaller open sector between them contains no line in $\mathcal{L}$. Let $\theta \in (0,\pi/2]$ denote the angle between these two lines.[/step]

[guided]The first simplification is to ignore signs. A root system is centrally symmetric: if $\alpha \in \Phi$, then $-\alpha \in \Phi$. Since the system is reduced, no other scalar multiple of $\alpha$ is a root. Thus each line through a root contributes exactly two roots, namely $\alpha$ and $-\alpha$. Formally, for each root $\alpha \in \Phi$ we define \begin{align*} L_\alpha := \mathbb{R}\alpha \subset V. \end{align*} The set of all root lines is \begin{align*} \mathcal{L} := \{L_\alpha : \alpha \in \Phi\}. \end{align*} Reducedness says that if $c\alpha \in \Phi$ for some $c \in \mathbb{R}$, then $c=\pm 1$. Therefore each line in $\mathcal{L}$ contains exactly two roots, one in each direction, and so \begin{align*} |\Phi| = 2|\mathcal{L}|. \end{align*} Since $V$ is two-dimensional, the lines through the origin form a circle with antipodal directions identified. We choose two adjacent root lines $L_\alpha$ and $L_\beta$, meaning that no other root line lies strictly in the smaller sector between them. The angle between lines is always taken in $(0,\pi/2]$, so we write this adjacent angle as $\theta \in (0,\pi/2]$.[/guided]

[step:Show that adjacent reflections force equally spaced root lines]For a root $\gamma \in \Phi$, let \begin{align*} s_\gamma: V &\to V \\ v &\mapsto v - 2\frac{(v,\gamma)}{(\gamma,\gamma)}\gamma \end{align*} be the reflection across the hyperplane $\gamma^\perp$. In dimension two, this reflection fixes the line $\gamma^\perp$ and sends every root line to another root line, because $s_\gamma(\Phi)=\Phi$. Equivalently, the reflection through the line $L_\gamma$ is $-s_\gamma$, and it also preserves the set $\mathcal{L}$ of root lines, since multiplying by $-1$ does not change a line. Therefore reflection through either adjacent root line $L_\alpha$ or $L_\beta$ preserves $\mathcal{L}$. Reflect $L_\beta$ through $L_\alpha$. The image is a root line making angle $\theta$ with $L_\alpha$ on the opposite side of $L_\alpha$. Similarly, reflecting $L_\alpha$ through $L_\beta$ produces a root line making angle $\theta$ with $L_\beta$ on the opposite side. Since there is no root line between $L_\alpha$ and $L_\beta$, these reflected lines are adjacent to $L_\alpha$ and $L_\beta$, respectively. Repeating this argument, the root lines occur at successive angular positions \begin{align*} 0,\theta,2\theta,\dots \end{align*} in the projective circle of total angular length $\pi$. Since $\mathcal{L}$ is finite, the process returns to the initial line after some positive integer $m$, and hence \begin{align*} m\theta = \pi. \end{align*} Thus \begin{align*} \theta = \frac{\pi}{m}, \end{align*} where $m=|\mathcal{L}|$.[/step]

[guided]The key point is that the root-system reflection axiom acts not only on roots but also on the lines spanned by roots. For $\gamma \in \Phi$, define \begin{align*} s_\gamma: V &\to V \\ v &\mapsto v - 2\frac{(v,\gamma)}{(\gamma,\gamma)}\gamma. \end{align*} The root-system axiom gives $s_\gamma(\Phi)=\Phi$. Hence, if $L_\delta$ is a root line, then \begin{align*} s_\gamma(L_\delta)=L_{s_\gamma(\delta)} \in \mathcal{L}. \end{align*} The map $s_\gamma$ is reflection across $\gamma^\perp$. In the plane, reflection through the line $L_\gamma$ is $-s_\gamma$. Multiplication by $-1$ fixes every line, so $-s_\gamma$ also preserves $\mathcal{L}$. Thus we may reflect root lines through root lines. Now take the adjacent pair $L_\alpha,L_\beta$. Reflect $L_\beta$ through $L_\alpha$. The image is again a root line, and elementary plane geometry shows that it lies on the opposite side of $L_\alpha$ at the same angle $\theta$. Because there was no root line between $L_\alpha$ and $L_\beta$, the reflected line is adjacent to $L_\alpha$ on the other side. Reflecting $L_\alpha$ through $L_\beta$ gives the next adjacent line beyond $L_\beta$. Repeating this reflection step produces root lines separated by the same angle $\theta$. Since the root system is finite, there are only finitely many root lines. The projective circle of unoriented lines has total angular length $\pi$, so after some positive integer $m$ steps the process returns to the original line: \begin{align*} m\theta = \pi. \end{align*} Therefore \begin{align*} \theta = \frac{\pi}{m}. \end{align*} The construction also shows that these $m$ lines are all the root lines, since an additional root line would have to lie in one of the adjacent sectors, contradicting the way the adjacent sectors were generated.[/guided]

[step:Use crystallographic integrality to restrict the adjacent angle]Choose roots $\alpha,\beta \in \Phi$ spanning the adjacent lines $L_\alpha,L_\beta$. Define the Cartan integers \begin{align*} p := 2\frac{(\beta,\alpha)}{(\alpha,\alpha)}, \qquad q := 2\frac{(\alpha,\beta)}{(\beta,\beta)}. \end{align*} The crystallographic condition gives $p,q \in \mathbb{Z}$. Their product is \begin{align*} pq &= 4\frac{(\alpha,\beta)^2}{(\alpha,\alpha)(\beta,\beta)} = 4\cos^2\theta. \end{align*} Since $0 < \theta \le \pi/2$, we have \begin{align*} 0 \le 4\cos^2\theta < 4. \end{align*} Thus $pq$ is an integer in $\{0,1,2,3\}$. Hence \begin{align*} 4\cos^2\theta \in \{0,1,2,3\}. \end{align*} The corresponding values of $\theta \in (0,\pi/2]$ are \begin{align*} \theta \in \left\{\frac{\pi}{2},\frac{\pi}{3},\frac{\pi}{4},\frac{\pi}{6}\right\}. \end{align*}[/step]

[guided]The crystallographic condition is what makes the classification finite. Choose roots $\alpha$ and $\beta$ lying on the adjacent lines. We define the two Cartan integers by \begin{align*} p := 2\frac{(\beta,\alpha)}{(\alpha,\alpha)}, \qquad q := 2\frac{(\alpha,\beta)}{(\beta,\beta)}. \end{align*} Because $\Phi$ is crystallographic, both $p$ and $q$ are integers. Multiplying these two integers eliminates the relative lengths of $\alpha$ and $\beta$: \begin{align*} pq &= \left(2\frac{(\beta,\alpha)}{(\alpha,\alpha)}\right) \left(2\frac{(\alpha,\beta)}{(\beta,\beta)}\right) \\ &= 4\frac{(\alpha,\beta)^2}{(\alpha,\alpha)(\beta,\beta)}. \end{align*} By the definition of the angle between the two lines, \begin{align*} \cos^2\theta = \frac{(\alpha,\beta)^2}{(\alpha,\alpha)(\beta,\beta)}. \end{align*} Therefore \begin{align*} pq = 4\cos^2\theta. \end{align*} Since $\theta \in (0,\pi/2]$, the number $4\cos^2\theta$ lies in the interval $[0,4)$. It is also an integer, because it equals $pq$. Hence \begin{align*} 4\cos^2\theta \in \{0,1,2,3\}. \end{align*} Solving these four possibilities gives \begin{align*} \theta = \frac{\pi}{2},\qquad \theta = \frac{\pi}{3},\qquad \theta = \frac{\pi}{4},\qquad \theta = \frac{\pi}{6}, \end{align*} respectively.[/guided]

[step:Count the root lines and roots in each possible angle] Since $\theta=\pi/m$ and \begin{align*} \theta \in \left\{\frac{\pi}{2},\frac{\pi}{3},\frac{\pi}{4},\frac{\pi}{6}\right\}, \end{align*} the number $m=|\mathcal{L}|$ of root lines is respectively \begin{align*} 2,\qquad 3,\qquad 4,\qquad 6. \end{align*} Because $\Phi$ is reduced, each root line contributes exactly the two roots $\pm \alpha$. Therefore \begin{align*} |\Phi| = 2m \in \{4,6,8,12\}. \end{align*} More precisely, the four cases give respectively \begin{align*} |\Phi|=4,\qquad |\Phi|=6,\qquad |\Phi|=8,\qquad |\Phi|=12. \end{align*} [/step]

[step:Identify the four angle cases with the four standard systems]It remains to identify the isomorphism type. Choose adjacent simple roots $\alpha,\beta \in \Phi$ so that $(\alpha,\beta)\le 0$. Define their Cartan integers by \begin{align*} a := 2\frac{(\beta,\alpha)}{(\alpha,\alpha)}, \qquad b := 2\frac{(\alpha,\beta)}{(\beta,\beta)}. \end{align*} Then $a,b \in \mathbb{Z}_{\le 0}$ and \begin{align*} ab = 4\cos^2\theta. \end{align*} The possible ordered pairs $(a,b)$, up to interchanging $\alpha$ and $\beta$, are therefore \begin{align*} (0,0),\qquad (-1,-1),\qquad (-1,-2),\qquad (-1,-3). \end{align*} These are exactly the Cartan data of \begin{align*} A_1\times A_1,\qquad A_2,\qquad B_2,\qquad G_2, \end{align*} respectively. The preceding steps show that every rank two reduced crystallographic root system has one of these four Cartan data, and the standard root systems listed above realize these four possibilities. Since a rank two reduced crystallographic root system is determined up to isomorphism by this rank two Cartan data, the four cases are mutually non-isomorphic and exhaustive. This proves the classification and the stated root counts.[/step]

[guided]After the angle and the number of lines have been determined, we still have to distinguish the possible relative lengths of adjacent roots. This information is recorded by the ordered Cartan integers. Choose adjacent roots $\alpha,\beta \in \Phi$ with $(\alpha,\beta)\le 0$, and define \begin{align*} a := 2\frac{(\beta,\alpha)}{(\alpha,\alpha)}, \qquad b := 2\frac{(\alpha,\beta)}{(\beta,\beta)}. \end{align*} The crystallographic condition gives $a,b \in \mathbb{Z}$. The sign convention $(\alpha,\beta)\le 0$ gives $a,b \le 0$. Their product is \begin{align*} ab = 4\cos^2\theta. \end{align*} The four possible values of $ab$ are $0,1,2,3$. Up to interchanging $\alpha$ and $\beta$, the corresponding non-positive integer pairs are \begin{align*} (0,0),\qquad (-1,-1),\qquad (-1,-2),\qquad (-1,-3). \end{align*} These pairs determine the rank two Cartan matrices \begin{align*} \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix}, \qquad \begin{pmatrix} 2 & -1 \\ -1 & 2 \end{pmatrix}, \qquad \begin{pmatrix} 2 & -2 \\ -1 & 2 \end{pmatrix}, \qquad \begin{pmatrix} 2 & -3 \\ -1 & 2 \end{pmatrix}, \end{align*} after possibly ordering the shorter and longer simple root in the last two cases. These are precisely the standard Cartan matrices of \begin{align*} A_1\times A_1,\qquad A_2,\qquad B_2,\qquad G_2. \end{align*} The standard systems realize the four configurations: two orthogonal root lines for $A_1\times A_1$, three equally spaced root lines for $A_2$, four for $B_2$, and six for $G_2$. Since the reflections generated by the two adjacent simple roots reconstruct the whole rank two system, this Cartan data determines the root system up to isomorphism. Therefore no further rank two reduced crystallographic root systems exist, and the four listed systems are pairwise non-isomorphic.[/guided]