= Solution
A continuous <Lorentz transformation> in one spatial dimension must preserve $t^2-x^2$. In the component connected to the identity, take the image of the time axis to have velocity $v$, with $|v|<1$, and write $B_v(1,0)^{\mathsf T}=\gamma(1,v)^{\mathsf T}$. Preservation of the <Minkowski norm> gives $\gamma^2(1-v^2)=1$, so $\gamma=(1-v^2)^{-1/2}$. The spacelike column must be orthogonal to the timelike column and have squared <Minkowski norm> $-1$; continuity fixes it to $\gamma(v,1)^{\mathsf T}$. Thus
$$
\boxed{x'=B_vx,\qquad
B_v=\frac1{\sqrt{1-v^2}}\begin{pmatrix}1&v\\v&1\end{pmatrix}.}
$$
Here $x=(t,x)^{\mathsf T}$ denotes the coordinate pair. This is the positive off-diagonal convention in the source; describing a passive frame moving at $+v$ instead reverses the sign of $v$.
Define the <rapidity> $\alpha=\operatorname{artanh}v$. The <hyperbolic function> identities give
$$
\cosh\alpha=\frac1{\sqrt{1-v^2}},\qquad
\sinh\alpha=\frac v{\sqrt{1-v^2}},\qquad
\boxed{B(\alpha)=\begin{pmatrix}\cosh\alpha&\sinh\alpha\\\sinh\alpha&\cosh\alpha\end{pmatrix}.}
$$
In <light-cone coordinates>, $t'+x'=e^\alpha(t+x)$ and $t'-x'=e^{-\alpha}(t-x)$. Therefore the derivatives transform oppositely:
$$
\partial_{t'}+\partial_{x'}=e^{-\alpha}(\partial_t+\partial_x),\qquad
\partial_{t'}-\partial_{x'}=e^{\alpha}(\partial_t-\partial_x).
$$
Use the field-transformation convention $\psi'(x')=S(\alpha)\psi(x)$. The component <action> in <3/c> is invariant under
$$
\boxed{\psi_R'(x')=e^{\alpha/2}\psi_R(x),\qquad
\psi_L'(x')=e^{-\alpha/2}\psi_L(x),\qquad
\Lambda_R(\alpha)=e^{\alpha/2},\quad\Lambda_L(\alpha)=e^{-\alpha/2}.}
$$
Indeed, the right-component <kinetic term> gains $e^{\alpha/2}e^{-\alpha}e^{\alpha/2}=1$, and the left-component <kinetic term> gains $e^{-\alpha/2}e^{\alpha}e^{-\alpha/2}=1$. The two factors in each <Dirac mass term> cancel as well. Finally $\det B=1$, so $dt'\,dx'=dt\,dx$ and the whole <action> is unchanged. These are the <chiral boost weights in two spacetime dimensions>.
In <spinor> notation, $S(\alpha)=\exp(\alpha\gamma^5/2)$ satisfies $S^{-1}\gamma^\mu S=B^\mu{}_{\nu}\gamma^\nu$ and $S^\dagger\gamma^0S=\gamma^0$, so $\bar\psi'=\bar\psi S^{-1}$. These identities directly verify covariance of the <Dirac equation> and the <Lorentz invariance> of its <action>. The matrices $S$ are not unitary with respect to the ordinary component norm; the <Dirac adjoint> supplies the invariant bilinear.
The statement about there being only <Lorentz boosts> applies to the identity component of the <One-dimensional Lorentz group>. The full <Lorentz group> also has discrete transformations: the concrete <spatial reflection> $\operatorname{diag}(1,-1)$ preserves the <Minkowski metric> but has <determinant> $-1$, so it cannot equal any displayed $B(\alpha)$, whose <determinant> is $+1$.
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