Since you're curious as to whether or not you're going in the right direction, allow me to address your approach first before giving you my solution (which follows a different approach).
I actually liked what you were trying to do and attempted to rescue it for a while. Then, I realized it might be a little too weak. I don't have a counterexample at hand but I'll share my intuition. You are trying to prove $\mathbb{E}\left\{ (S_{n+1}^2 - s_{n+1}^2) \mathbb{1}_A \right\} \geq 0$, or intuitively that conditional on the partial sums of $S_n$ being small (i.e. event $A$ occurring), your realized variance $S_{n+1}^2$ is in expectation at least as large as the expected (unconditional) variance $s_{n+1}^2$. I just can't imagine why small partial sums would give you large variance. The situation is usually reversed: having large partial sums allows you to say the variance is large (Chebyshev, Kolmogorov, Doob inequalities all work in this context). I don't think there is enough data in the problem to prove a converse inequality.
On to the solution.
First of all, consider the slightly more natural stopping time $\tau := n \wedge \inf \{ m \geq 1 : |S_m| > x \}$. This is different from your $N$ in a couple of ways that make it easier to use. Furthermore, it is clearly a bounded stopping time and such that $S_\tau^2 \leq (x+K)^2$.
In view of $S_n^2 - s_n^2$ being a martingale and $\tau$ a bounded stopping time, the optional stopping theorem (or the variant thereof you refer to as Theorem 5.4.1) gives $\mathbb{E} S_\tau^2 = \mathbb{E} s_\tau^2$. In view of $S_\tau^2 \leq (x+K)^2$, we conclude that
$$ (x+K)^2 \geq \mathbb{E} S_\tau^2 = \mathbb{E} s_\tau^2 = \mathbb{E} \left\{ \mathbb{1}_A s_\tau^2 \right\} + \mathbb{E} \left\{ \mathbb{1}_{A^c} s_\tau^2 \right\} \geq \mathbb{E} \left\{ \mathbb{1}_A s_\tau^2 \right\}$$
where $A = \cap_{1 \leq m \leq n} \{ |S_m| \leq x \}$, as with your approach. Notice that when $A$ occurs we have $\tau = n$ and therefore $s_\tau^2 = s_n^2$, the latter being a constant, so
$$ (x+K)^2 \geq \mathbb{E} \left\{ \mathbb{1}_A s_\tau^2 \right\} = \mathbb{E} \left\{ \mathbb{1}_A s_n^2 \right\} = \mathbb{E} \mathbb{1}_A \cdot s_n^2 = \mathbb{P}(A) \mathbb{E} S_n^2 $$
which is the required result.
As you observed, it suffices to check that $\mathbb E\left[M_n\mathbf{1}_{n\geqslant T}\right]=0$, since $M_n\geqslant 0$ will imply that $M_n\mathbb{1}_{n\geqslant T}=0$ almost surely.
To do so, we decompose $\mathbf{1}_{n\geqslant T}$ as the sum of $\mathbf{1}_{T=k}$ in order to get
$$
\mathbb E\left[M_n\mathbf{1}_{n\geqslant T}\right]=\sum_{k=0}^n\mathbb E\left[M_n\mathbf{1}_{T=k}\right]
$$
and to reduce the problem to show that $\mathbb E\left[M_n\mathbf{1}_{T=k}\right]=0$ for each $n\geqslant k$.
Denote by $\left(\mathcal F_n\right)_{n\geqslant -1}$ the filtration associated to the martingale $\left(M_n\right)_{n\geqslant 0}$.
We notice that $\{T=k\}$ belongs to $\mathcal F_k$, as $\{T=k\}=\{M_k=0\}\cap\bigcap_{j=0}^{k-1}\{M_{j}>0\}$ if
$k\geqslant 1$ and $\{T=0\}=\{M_0=0\}$. As a consequence,
$$
\mathbb E\left[M_n\mathbf{1}_{T=k}\right]=\mathbb E\left[\mathbb{E}\left[M_n\mathbf{1}_{T=k}\mid\mathcal F_k\right]\right]= \mathbb E\left[\mathbf{1}_{T=k}\mathbb{E}\left[M_n\mid\mathcal F_k\right]\right]
$$
and using the martingale property [until here, we only used $\mathcal F_n$-measurability of $M_n$], we get
$$
\mathbb E\left[M_n\mathbf{1}_{T=k}\right]= \mathbb E\left[\mathbf{1}_{T=k}M_k\right]
$$
which is $0$ because $\mathbf{1}_{T=k}M_k=0$ almost surely.
Best Answer
From the definition of a martingale, we want to show that $\mathbb{E}[X_t | \mathcal F_s] = X_s$ for all $s < t$, which is equivalent to showing that for all $A \in \mathcal F_s$ we have $\mathbb{E}[X_t 1_A] = \mathbb{E}[X_s 1_A]$. We define a stopping time $\tau := t 1_A + s 1_{A^c}$ (check this is a stopping time!), and from the hypotheses we have
\begin{align*} \mathbb{E}[X_s] = \mathbb{E}[X_0] &= \mathbb{E}[X_\tau] = \mathbb{E}[X_t 1_A + X_s 1_{A^c}] \end{align*}
so $\mathbb{E}[X_s 1_A] = \mathbb{E}[X_t 1_A]$. We conclude $\mathbb{E}[X_t | \mathcal F_s] = X_s$ and therefore $(X_n)$ is a martingale since we already had that it was adapted and integrable.