Let $t\leq t_{1}$ be an arbitrary time at which the solution $P(t)$ of the following Riccati differential equation exists:
$\dot{P}(t) =-P(t) A(t) - A^{T}(t)P(t) -Q(t) + P(t)B(t)R^{-1}(t)B^{T}(t)P(t)\\ P(t_{1})=M$
with $M=M^{T}\geq 0$, $Q(t)=Q^{T}(t)\geq 0$ and $R(t)=R^{T}>0$ for all $t\in [t_{0},t_{1}]$. $Q,R,M,A,B$ are real and continuous matrices of appropriate dimensions. How can I show that $P(t)$ is a symmetric matrix and $P(t)\geq 0$.
I am assuming that you meant to have $-Q(t)$ in the problem statement.
In general, if $C$ is a closed subspace and $f(x,t) \in C$ for all $x \in C$ (and all $t \ge t_1$) then if $x_0 \in C$ we have $x(t) \in C$ for all $t \ge t_1$.
This can be proved using the Picard iteration applied to the operator $S$ defined by $(Sx)(t) = x_0 + \int_{t_1}^t f(x(\tau),\tau) d \tau$.
In the above, take $C$ to be the space of symmetric matrices and show that the right hand side is closed in the above sense.
Alternative:
A much simpler solution would be to use uniqueness of solution. If $P(t_1)$ is symmetric, then note that $t \mapsto P^T(t)$ is a solution to the differential equation with the same initial condition, hence $P(t) = P^T(t)$ for all $t \ge t_1$.