The simple Markov property says that if $X$ is $Markov(\lambda, P)$, then conditional on $X_m = i$, $(X_{m+n})_{n\geq 0}$ is $Markov(\delta_i, P)$ and is independent of the random variables $X_0, ..., X_m$.
I have proved the first part of the theorem, that $(X_{m+n})_{n\geq 0}$ is $Markov(\delta_i, P)$ using the law of total probability. However, I'm not sure how to prove independence. It suffices to show that $$ P(X_{i_1} = x_1, ..., X_{i_k} = x_k, X_0=x_0', ..., X_m = x_m' | X_m = i) = P(X_{i_1} = x_1, ..., X_{i_k}=x_k|X_m=i)P(X_0=x_0', ..., X_m = x_m' | X_m = i) $$ but I'm not sure how. I think that the property of $Markov(\lambda, P)$ chains that $$P(X_0=x_0, \dots, X_n = x_n) = \lambda(x_0) P(x_0, x_1) \cdots P(x_{n-1}, x_n)$$ may help, but I can't apply this directly.
I suspect the derivation involves summing over the possible values for the intermediary random variables (using the law of total probability), but I can't quite get the algebra to follow.
Indeed, one can't apply the property directly, but by explicitly introducing the intermediary variables and summing over using the law of total probability, the desired result follows from a long (though routine) calculation.