Assume $X_1, \ldots, X_n$ are independent and identically distributed random variables where $X \in \{-1, 1\}$ with equal probabilities, and $S_k$ represents their cumulative sum of up to $k$ variables.
The exercise is to show that $\mathbb{E}S_{\tau}^2 = \mathbb{E}\tau$, where $\tau$ is a $(\mathcal{F})_n$ measurable stopping time.
On one hand, it feels like
$$ \mathbb{E}\tau = \sum_{j=1}^nj \mathbb{P}(\tau=j)=\sum_{j=1}^n j \mathbb{E}(\mathbb{1}_{\tau = j})$$
On the other hand,
$$ \mathbb{E}(S^2_\tau)=\sum_{j=1}^n\mathbb{E}(S_j^2 \mathbb{1}_{\tau = j})=\sum_{j=1}^n\mathbb{E}( \mathbb{E}[S_j^2 \mathbb{1}_{\tau = j}|\mathcal{F}_j]) = \sum_{j=1}^n\mathbb{E}(\mathbb{1}_{\tau = j} \mathbb{E}[S_j^2 |\mathcal{F}_j])$$
In essence, it is easy to check that $\mathbb{E}(S_j^2)=j$ due to equal probabilities.
My question: is the last equation a proper way to pull out the $j$ from the expectation? That is, by stating that $\mathbb{E}(S_j^2 | \mathcal{F}_j)=j$? It feels like double expectation here is one way to pull out the $j$, but not entirely sure whether $\mathcal{F}_j$ is the right choice here, since it makes the $S_j^2$ measurable.
On the other hand, is pulling out the $j$ necessary, or is it for some reason obvious that $$ \mathbb{E}S_j^2 \mathbb{1}_{\tau=j} = j \mathbb{P}(\tau = j)?$$
First, $\{S_k^2-k\}$ is a martingale. Since $\tau\le n$, the optional stopping theorem implies $$ \mathsf{E}[S_{\tau}^2-\tau]=\mathsf{E}[S_1^2-1]=0. $$
Regarding your question, let $n=4$, $\tau:=\min\left(\{k\ge 1:S_k\ge 1 \}\cup\{n\}\right)$. Then $$ 3\times\mathsf{P}(\tau=3)=\frac{3}{8}, $$ but $$ \mathsf{E}[S_\tau^21\{\tau=3\}]=\mathsf{E}[S_\tau^2\mid \tau=3]\mathsf{P}(\tau=3)=\frac{1}{8}. $$
Direct computation shows that $$ \mathsf{E}\tau=\sum_{j=1}^4 j\mathsf{P}(\tau=j)= 1\times\frac{1}{2}+2\times 0+3\times\frac{1}{8}+4\times\frac{3}{8}=\frac{19}{8} $$ and $$ \mathsf{E}S_\tau^2=\sum_{j=1}^4 \mathsf{E}[S_\tau^21\{\tau=j\}]=\frac{1}{2}+0+\frac{1}{8}+\frac{7}{4}=\frac{19}{8}. $$