Let the distance from $C$ to $P$ be $a$, and let the $AC = BC = \frac{\sqrt{2}}{2} AB = l$. We have $AP = \sqrt{a^2 + l^2}$, and since $CR$ is the altitude from $C$ it is by area relations $\frac{la}{\sqrt{a^2 + l^2}}$. By Pythagorean again, $AR = \sqrt{l^2 - \frac{l^2 a^2}{l^2 + a^2}}$. Since $SA = CR$, we get \begin{align*} RS &= \sqrt{l^2 - \frac{l^2 a^2}{l^2 + a^2}} - \frac{la}{\sqrt{a^2 + l^2}} \\ &= \frac{1}{\sqrt{a^2 + l^2}} \sqrt{a^2 l^2 + l^4 - l^2 a^2} - \frac{la}{\sqrt{a^2 + l^2}} \\ &= \frac{l(l-a)}{\sqrt{l^2 + a^2}} \end{align*}The first part is now done. Note that $\angle CAP = arctan(\frac{a}{l})$. Thus, we have $\angle BAP = 45^{\circ} - arctan(\frac{a}{l})$. Using the law of cosines, \begin{align*} SQ &= \sqrt{\frac{l^2}{2} + \frac{l^2 a^2}{a^2 + l^2} - 2 \frac{la}{\sqrt{l^2 + a^2}} \cdot l \frac{\sqrt{2}}{2} \cdot (sin(45^{\circ}) sin(\angle BAP) + cos(45^{\circ}) cos(\angle BAP))} \\ &= \sqrt{\frac{l^2}{2} + \frac{l^2 a^2}{a^2 + l^2} - 2 \frac{la}{\sqrt{l^2 + a^2}} \cdot l \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{2}}{2} (\frac{a + l}{\sqrt{a^2 + l^2}})} \\ &= \frac{1}{\sqrt{a^2 + l^2}} \sqrt{\frac{a^2 l^2}{2} + \frac{l^4}{2} + l^2 a^2 - l^2 a^2 - l^3 a} \\ &= \frac{l}{\sqrt{2} \sqrt{a^2 + l^2}} \sqrt{a^2 - 2al + l^2} \\ &= \frac{l (l-a)}{\sqrt{2} \sqrt{a^2 + l^2}} = \frac{1}{\sqrt{2}} RS \end{align*}as desired. $\square$