Prove Why $B^2 = A$ exists?

The theory reasons why are given above. To prove this however one simply needs to find a matrix such that $B^2=A$ in this case

$B=$ \begin{array}{cccccccc} \frac{1}{90} \left(-\frac{75}{\sqrt{2}}-70 \sqrt{2}+181 \sqrt{5}\right) & -\frac{2}{\sqrt{5}} & \frac{1}{180} \left(-\frac{75}{\sqrt{2}}-70 \sqrt{2}+91 \sqrt{5}\right) & \frac{1}{\sqrt{5}} & \frac{1}{120} \left(-\frac{75}{\sqrt{2}}-130 \sqrt{2}+73 \sqrt{5}\right) & \frac{4}{9} \left(\frac{3}{\sqrt{2}}+10 \sqrt{2}-\frac{17 \sqrt{5}}{2}\right) & \frac{4}{45} \left(\frac{15}{\sqrt{2}}+20 \sqrt{2}-\frac{43 \sqrt{5}}{2}\right) & \frac{1}{2 \sqrt{5}} \\ \frac{15}{16 \sqrt{2}} & \sqrt{2} & \frac{15}{32 \sqrt{2}} & 0 & \frac{29}{64 \sqrt{2}} & 0 & 0 & 0 \\ \frac{1}{90} \left(\frac{285}{\sqrt{2}}+320 \sqrt{2}-362 \sqrt{5}\right) & \frac{4}{\sqrt{5}} & \frac{1}{180} \left(\frac{285}{\sqrt{2}}+320 \sqrt{2}-182 \sqrt{5}\right) & -\frac{2}{\sqrt{5}} & \frac{1}{120} \left(\frac{285}{\sqrt{2}}+260 \sqrt{2}-146 \sqrt{5}\right) & \frac{1}{9} (-8) \left(\frac{3}{\sqrt{2}}+10 \sqrt{2}-\frac{17 \sqrt{5}}{2}\right) & \frac{1}{45} (-8) \left(\frac{15}{\sqrt{2}}+20 \sqrt{2}-\frac{43 \sqrt{5}}{2}\right) & -\frac{1}{\sqrt{5}} \\ \frac{1}{12} \left(\frac{21}{2 \sqrt{2}}-8 \sqrt{2}+8 \sqrt{5}\right) & 2 \left(\sqrt{2}-\sqrt{5}\right) & \frac{1}{24} \left(\frac{21}{2 \sqrt{2}}-8 \sqrt{2}+8 \sqrt{5}\right) & \sqrt{5} & \frac{1}{16} \left(\frac{5}{2 \sqrt{2}}+40 \sqrt{2}-40 \sqrt{5}\right) & \frac{1}{3} (-2) \left(\frac{3}{\sqrt{2}}+5 \sqrt{2}-5 \sqrt{5}\right) & \frac{1}{3} (-2) \left(\frac{3}{\sqrt{2}}-\sqrt{2}+\sqrt{5}\right) & 0 \\ -\frac{1}{\sqrt{2}} & 0 & -\frac{1}{2 \sqrt{2}} & 0 & \frac{5}{4 \sqrt{2}} & 0 & 0 & 0 \\ -\frac{1}{2 \sqrt{2}} & 0 & -\frac{1}{4 \sqrt{2}} & 0 & -\frac{3}{8 \sqrt{2}} & \frac{5}{2 \sqrt{2}} & \frac{1}{2 \sqrt{2}} & 0 \\ \frac{1}{2 \sqrt{2}} & 0 & \frac{1}{4 \sqrt{2}} & 0 & \frac{3}{8 \sqrt{2}} & -\frac{1}{2 \sqrt{2}} & \frac{3}{2 \sqrt{2}} & 0 \\ \frac{1}{2} \left(-\frac{1}{\sqrt{2}}-2 \sqrt{2}+2 \sqrt{5}\right) & 0 & \frac{1}{4} \left(-\frac{1}{\sqrt{2}}-2 \sqrt{2}+2 \sqrt{5}\right) & 0 & \frac{1}{8} \left(-\frac{3}{\sqrt{2}}-2 \sqrt{2}+2 \sqrt{5}\right) & 0 & 0 & \sqrt{5} \\ \end{array}

I borrow (and slightly modify) an explanation that can be found in the excellent thread https://mathoverflow.net/questions/14106/finding-the-square-root-of-a-non-diagonalizable-positive-matrix

$f$ being any function, sufficient derivable, one can write, for example for a $4 \times 4$ Jordan block:

$$f\left( \left[ \begin{array} [c]{cccc}% \lambda & 1 & & \\ & \lambda & 1 & \\ & & \lambda & 1\\ & & & \lambda \end{array} \right] \right) =\left[ \begin{array} [c]{cccc}% f\left( \lambda\right) & f^{\prime}\left( \lambda\right) & \frac{1} {2!}f^{\prime\prime}\left( \lambda\right) & \frac{1}{3!}f^{\prime \prime\prime}\left( \lambda\right) \\ & f\left( \lambda\right) & f^{\prime}\left( \lambda\right) & \frac {1}{2!}f^{\prime\prime}\left( \lambda\right) \\ & & f\left( \lambda\right) & f^{\prime}\left( \lambda\right) \\ & & & f\left( \lambda\right) \end{array} \right]$$

(it is based on a certain Taylor expansion of $I+N$ where $N$ is a nilpotent matrix).

It suffices now to take $f(x)=\sqrt{x}$ to get for the square root of a Jordan block of similar form as above but $3 \times 3$:

$$\left[ \begin{array} [c]{ccc}% \sqrt{\lambda} & \dfrac{1}{2\sqrt{\lambda}} & -\dfrac{1}{8 (\lambda)^{3/2}} \\ & \sqrt{\lambda} & \dfrac{1}{2\sqrt{\lambda}} \\ & & \sqrt{\lambda} \end{array} \right]$$

By putting $A$ into Jordan normal form you have found a non-singular matrix $P$ such that $$ A = P^{-1}CP $$ where $C$ is your matrix shown above. It is easy to find a matrix sqare root of $C$, for example, in the upper left start with $\pmatrix{\sqrt{2} &0\\\frac14\sqrt{2}& \sqrt{2}}$

Then $$(P^{-1}DP)^2 = P^{-1}D(PP^{-1})DP = P^{-1}D^2P=P^{-1}CP=A $$