Dissolution of CO2 in water
Assumptions:

Reaction of carbon dioxide with water is neglected.
Vapour pressure of water is negligible.
Volume of solution does not change on dissolution of carbon dioxide.
Moles of carbon dioxide dissolved in water is very less as compared to moles of water and thus, $X_{\ce{CO2_{(aq.)}}}≈\frac{n_{\ce{CO2(aq)}}}{n_{\ce{H2O(l)}}}$


Initially no $\ce{CO2}$ is dissolved in water and so initial moles of $\ce{H2O_{(l)}}$ and $\ce{CO2_{(g)}}$ can be calculated as follows:
$$
n_{{\ce{H2O_{(l)}}}_i} = \frac{\rho V}{M} =\pu{\frac{500 \times 0.9975}{18} mol}= \pu{27.71 mol}\\
n_{{\ce{CO2_{(g)}}}_i} = \frac{P_iV}{RT} = \pu{ \frac{50 \times 0.07}{0.0821 \times 283} mol}= \pu{0.15 mol}\\
$$
Now after the equilibrium is settled some amount of carbon dioxide will dissolve in water according to Henry's law.
Final pressure in the container can be written as:
$$
P_f= \frac{n_{{\ce{CO2_{(g)}}}_f}RT}{V}\\
$$
And by Henry's law we can say,
$$
P_f= K_H \times X_{\ce{CO2_{(aq.)}}}
$$
Thus from both these equations we can conclude:
$$
K_H \times  \frac{n_{\ce{CO2(aq)}}}{n_{\ce{H2O(l)}}} = \frac{n_{{\ce{CO2_{(g)}}}_f}RT}{V}\\
$$
We also known that,
$$
n_{{\ce{CO2_{(g)}}}_f} = n_{{\ce{CO2_{(g)}}}_i} - n_{{\ce{CO2_{(aq)}}}_f} \\
$$
So,
$$
K_H \times  \frac{n_{\ce{CO2(aq)}_f}}{n_{\ce{H2O(l)}}} = \frac{(n_{{\ce{CO2_{(g)}}}_i} - n_{{\ce{CO2_{(aq)}}}_f})RT}{V}\\
$$
Solving this we get,
$$
n_{\ce{CO2(aq)}_f} = \pu{0.135 mol}\\
$$
This correspondingly means:
$$
n_{{\ce{CO2_{(g)}}}_f}= \pu{0.015 mol}\\
P_f= \pu{5 atm}\\
$$
The final pressure is less than initial pressure as expected. The mistake you were making was implicitly considering $P_f=P_i$ when finding the mole fraction of carbon dioxide in water, which clearly isn't the case.