$\min_{\alpha \in \mathbb{R}^s} ||b-R\alpha ||_2 = \min_{\alpha \in \mathbb{R}^s} ||b-AS\alpha ||_2$
$\begin{array}{ll}
-& = \min_{x \in Vect\left(x_0, x_1, \hdots, x_{k-1} \right)} ||b-AS\alpha ||_2\\
-& \leqslant \min_{x \in Vect\left( S_{k-1} \right)} ||b-Ax ||_2\\
+& = \min_{x \in Vect\left(S_{k-s}, S_{k-s+1}, \hdots, S_{k-1} \right)} ||b-AS\alpha ||_2\\
+& = \min_{x \in Vect\left(x_{k-s}, x_{k-s}+1, \hdots, x_{k-1} \right)} ||b-AS\alpha ||_2\\
+& \leqslant \min_{x \in Vect\left( x_{k-1} \right)} ||b-Ax ||_2\\
& \leqslant ||b-Ax_{k-1}||
\end{array}$
\end{proof}