Coax Cable Wave Impedance, Capacitive Load
why does the Real Component of the Impedance fall down to 100 Ohm
at frequencies above 10 MHz ?

A transmission line's characteristic impedance (\$Z_0\$) is determined by this equation: -
$$Z_0 = \sqrt{\dfrac{R + j\omega L}{G + j\omega C}}$$
\$\color{red}{\text{This is mathematically set in stone.}}\$ See the derivation here. Also related to this question asked previously by the OP.
As frequency rises higher, the two components \$j\omega L\$ and \$ j\omega C\$ dominate over R and G hence, at higher frequencies the formula becomes: -
$$Z_0 = \sqrt{\dfrac{j\omega L}{j\omega C}} = \sqrt{\dfrac{L}{C}}$$
And, with your values of L = 250 nF and C = 50 pF you get: -
$$Z_0 = \text{50 ohms}$$
Now, if you terminate a transmission line with 50 ohms you provide the perfect matching situation where the input impedance to the coax is also 50 ohms. This may sound strange but, if the transmission line were not terminated in 50 ohms and was infinite in length you would still see 50 ohms.
So, the 50 ohms of the transmission line plus the 50 ohms from R1 produce an overall input impedance of 100 ohms.