= Solution
The <approximate source bound for coercive Tikhonov regularization> first allows a fixed sufficiently large $r$ to make $\eta_r$ small, then a sufficiently small $\alpha$ to make $\sqrt\alpha\beta r$ small. This order of limits proves $R_\alpha f\to K^\dagger f$ for every datum in its domain.
For the noise error let $v=R_\alpha g$. Its <normal equation for coercive Tikhonov regularization> gives
$$
\|Kv\|^2+\alpha\|Bv\|^2=\langle g,Kv\rangle\leq\|g\|\|Kv\|.
$$
Completing the square shows $\alpha\beta^2\|v\|^2\leq\|g\|^2/4$, hence $\|R_\alpha\|\leq1/(2\beta\sqrt\alpha)$. The <noise-bias decomposition for linear regularization> now gives
$$
\|R_\alpha f^\delta-K^\dagger f\|\leq\frac{\delta}{2\beta\sqrt\alpha}+\|R_\alpha f-K^\dagger f\|.
$$
Therefore the sufficient <regularization parameter choice> is
$$
\boxed{\alpha(\delta)\to0,\qquad\delta^2/\alpha(\delta)\to0.}
$$
For example $\alpha(\delta)=\delta^q$ with $0<q<2$ works. Coercivity makes $K^*K+\alpha B^*B$ boundedly invertible for each positive $\alpha$, so these are continuous linear reconstruction operators and form a <convergent regularization of an inverse problem>.
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