= Solution
Relative to a <filtration> $(\mathcal F_n)$, a <martingale> is an adapted integrable process with $\mathbb E[M_{n+1}\mid\mathcal F_n]=M_n$; a <supermartingale> has the corresponding inequality $\leq$. A <stopping time> $T$ has $\{T\leq n\}\in\mathcal F_n$ for every $n$.
For $T\leq N$, the pathwise identity
$$
M_T=M_0+\sum_{k=1}^N\mathbf1_{\{T\geq k\}}(M_k-M_{k-1})
$$
has $\mathcal F_{k-1}$-measurable indicators. Conditioning each summand gives zero expected increment for a <martingale>, and a nonpositive one for a <supermartingale>. All terms are integrable because the sum is finite. Thus $\mathbb EM_T=\mathbb EM_0$ and $\mathbb E\hat M_T\leq\mathbb E\hat M_0$, proving the requested inequality when their initial values agree.
A <self-financing strategy> holds $\pi_n$ shares during period $n$, chosen using information at time $n-1$. Its cash investment is $X_{n-1}-\pi_nS_{n-1}$, which earns the risk-free return. Hence
$$
X_n=\pi_nS_n+(1+r)(X_{n-1}-\pi_nS_{n-1})
=(1+r)X_{n-1}+\pi_n[S_n-(1+r)S_{n-1}].
$$
Use the terminal condition $V(N,x,s)=U(x)$ from the original PDF. Conditional on the current state and past, the next return has the law of $\xi_1$ and is independent of the past. Therefore the conditional expectation of $V(n+1,X_{n+1},S_{n+1})$ for the chosen holding $\pi_{n+1}$ is at most its supremum over all holdings, namely $V(n,X_n,S_n)$. Assuming the finite values and integrability implicit in this stochastic-control formulation, the value process is a <supermartingale>.
It follows that $\mathbb E U(X_N)\leq V(0,X_0,S_0)$ for every admissible strategy. If the strategy $\pi^*$ makes this process a <martingale>, equality holds for it. \b[$\pi^*$ attains the upper bound and is optimal.] No existence of an optimizer or finiteness of an unconstrained supremum is being assumed without the stated admissibility conditions.
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