I think I have found an answer and please correct me if there are any other reasons behind this justification.

Since $x^{*}(t)=\{x(t)\in\Omega|J(x)<\min J(y), \forall { y \in \Omega}\}$ is optimal between all admissible trajectories $x(t)\in\Omega$ for $t\in[t_0,t_1]$ with $x^{*}(t_0)=x(t_0)=x_0$ and $x^{*}(t_1)\neq{x(t_1)}$, it is still optimal for the fixed end point problem case $\Big((t_0,x_0)$ and $(t_1,x^{*}(t_1))\Big)$ as well and therfore the necessary conditions that hold for $x^{*}(t)$ in fixed end point case will hold in this case. Since, Euler-Lagrange condition was a necessary condition for $x^{*}(t)$ being optimal in fixed end point case, Eq. (2) holds true for this case and substituting this result in Eq. (3), the other condition in Eq. (5) can be found easily.

I think I have found an answer and please correct me if there are any other reasons apart from the following justification.

Since $x^{*}(t)=\{x(t)\in\Omega|J(x)<\min J(y), \forall { y \in \Omega}\}$ is optimal between all admissible trajectories $x(t)\in\Omega$ for $t\in[t_0,t_1]$ with $x^{*}(t_0)=x(t_0)=x_0$ and $x^{*}(t_1)\neq{x(t_1)}$, it is still optimal for the fixed end point problem case $\Big((t_0,x_0)$ and $(t_1,x^{*}(t_1))\Big)$ as well and therfore the necessary conditions that hold for $x^{*}(t)$ in fixed end point case will hold in this case. Since, Euler-Lagrange condition was a necessary condition for $x^{*}(t)$ being optimal in fixed end point case, Eq. (2) holds true for this case and substituting this result in Eq. (3), the other condition in Eq. (5) can be found easily.