$\newcommand{\ee}{\mathbb{E}}$The fact that for all $n$ we have $\ee[S_n \mid \mathcal{F_{n-1}}]=S_{n-1} ~\text{a.s.}$ and $\ee[ |S_n|]<\infty $ is usually shown explicitly when showing something is a martingale. I have not really seen the explicit calculation for adaptedness of $S_n$ and I don't know how to do it.
Given $S_n$ is a SSRW, would this be sufficient in showing that $S_n$ is adapted?
Take any $B \in \mathcal{B}(\mathbb{R})$ then the inverse image
$$ \begin{equation} \begin{split} S_N^{-1}(B)~=&~\{\omega: X_1(\omega)+\dots +X_N(\omega) \in B\} \\~=&~\{ X_1(\omega) \in B_1\} \cup\cdots\cup\{ X_N(\omega) \in B_N \} \end{split} \end{equation} $$
Then since $\{X_n\}_n^N$ are random variables and measurable with respect to $\mathcal{F}_N$. What would each $B_k$ be, for fixed $\omega$ a set in which all of the elements are shifted by the values of the other $X_k$? something like: $$B_k = B-\sum_{\substack{n=1\\n \neq k}}^N X_n(\omega)$$
Or $\sum_{k=1}^N B_k \subset B$
Given a positive integer $n$, we have $$S_n = \sum_{i=1}^n X_i$$ where $X_i$ is $\mathcal F_i$-measurable. Since $i<j$ implies $\mathcal F_i\subset\mathcal F_j$, each $X_i$ is $\mathcal F_n$-measurable. Now, given $a\in\mathbb R$, let $\{q_n\}$ be an enumeration of $\mathbb Q\cap(-\infty,a]$, then \begin{align} (X_1+X_2)^{-1}((-\infty,a]) &= \{\omega: X_1(\omega) + X_2(\omega)\leqslant a\}\\ &= \bigcup_{n=1}^\infty\{\omega:X_1(\omega)\leqslant q_n\}\cap\{\omega: X_2(\omega)\leqslant a-q_n\}\\ &= \bigcup_{n=1}^\infty X^{-1}_1((-\infty,q_n])\cap X^{-1}_2((-\infty,a-q_n]), \end{align} so that $X_1+X_2$ is $\mathcal F_2$-measurable. By induction it follows that $S_n$ is $\mathcal F_n$-measurable and hence the process $\{S_n\}$ is adapted with respect to the filtration $\{\mathcal F_n\}$.