Let $S_n$ be a simple symmetric random walk on the positive and negative integers, with $S_0 = 0$. Show whether each of the following is a stopping time: $$ U = \min\{n \geq 5 : S_n = S_{n-5} + 5\} $$
and $$ V = U - 5 $$
My solution(EDIT):
First of all noticing that it's a min, this is suggesting that U is a stopping time. $$ (U \leq k) \! \begin{aligned}[t] & = \bigcup^{k}_{n=5} \{S_n = S_{n-5} + 5\} \\ & = \bigcup \{S_n \leq S_{n-5} + 5\} \cup \{S_n \leq S_{n-4} + 5\}^c \in F_k \end{aligned} $$ Is that correct?
About $V$:
$U = V + 5$, so we need $5$ more times of information in order to have a stopping time and this is not possible. How could this be mathematically being proven?
You wrote (in the 1st version of the question) $(U \leq n) = \bigcup^{\infty}_{5} \{S_n = S_{n-5} + 5\}$. It looks like $(U \leq n) = \bigcup^{\infty}_{n = 5} \{S_n = S_{n-5} + 5\}$. L.h.s. depends on $n$, R.h.s. doesn't depend on $n$. Hence the solution if false.
Hint for the first part: $(U \leq k) = \bigcup^{k}_{n = 5} \{S_n = S_{n-5} + 5\}$.
Solution for the second part: $(U -5 \leq k) = (U \leq k+5)= \bigcup^{k+5}_{n = 5} \{S_n = S_{n-5} + 5\}$. Put $j = n-5$. We have $(U -5 \leq k) = \bigcup^{k+5}_{n = 5} \{S_n = S_{n-5} + 5\} = \bigcup^{k}_{j = 0} \{S_{j+5} = S_{j} + 5\} \notin F_k$ if $S_k$ is a sum of independent Rademacher r.v., $k=0$ and $F_k$ is a natural filtration of $\{ S_k, k \ge 0 \}$. Indeed, if $k = 0$ then an event $(U -5 \leq k) = \bigcup^{0}_{j = 0} \{S_{j+5} = S_{j} + 5\} = \{S_5 = S_0 + 5 \} = \{ S_5 = 5 \}$ have probability $p \in (0,1)$ and $F_0 = \{ \varnothing, \Omega \}$ hence $\{(U - 5) \le 0 \} \notin F_0$.
Addition about the second version of the OP's solution. It looks like you implied that $\{S_n = S_{n-5} + 5\} = \{S_n \leq S_{n-4} + 5\} \cup \{S_n < S_{n-5} + 5\}^c$ but it's not true. Suppose that $S_n = 10$, $S_{n-4}=6$, $S_{n-5}=7$. Hence $S_n \ne S_{n-5} + 5$ but $S_n \le S_{n-4} + 5$. So the second version of solution is not true.
Addition about the third version of the OP's solution. It looks like you implied that if $n \le k$ then it's obvious that $\{S_n \leq S_{n-5} + 5\} \cup \{S_n \leq S_{n-4} + 5\}^c \in F_k$, but it's not obvious that $\{S_n = S_{n-5} + 5\} \in F_k$. I don't agree with it. It's more obvious that $\{S_n = S_{n-5} + 5\} \in F_k$ than that $\{S_n \leq S_{n-5} + 5\} \cup \{S_n \leq S_{n-4} + 5\}^c \in F_k$.
Indeed, $S_n$ in $F_n$ measurable and $n \le k$, hence $S_n$ is $F_k$ measurable. Further, $S_{n-5}$ in $F_{n-5}$ measurable and $n-5 \le n \le k$, hence $S_{n-5}$ is $F_k$ measurable. Hence $S_n$ and $S_{n-5}$ are $F_k$ measurable and thus $S_n-S_{n-5}-5$ is $F_k$ measurable. Thus $\{S_n - S_{n-5} - 5 = 0 \} \in F_k$ and it follows that $\{S_n = S_{n-5} + 5\} \in F_k$. It's much more obvious than the relation $\{S_n \leq S_{n-5} + 5\} \cup \{S_n \leq S_{n-4} + 5\}^c \in F_k$.
Maybe you supposed that $\{S_n \le \xi \} \in F_n$ for all $\xi$ or smth.like that, but it's not true.