We know that in a Banach space $X$, a subset $U$ of $X$ is closed iff the limit of every convergent sequence in $U$ is also in $U$. Therefore, if all of the sequences in $U$ are not convergent, then $U$ is closed. Is that true?
Actually, I am trying to prove $U=\{(1+1/n)e_n\}$ is closed, because every sequence in $U$ is divergent, where $\{e_n\}$ is an orthonormal sequence in an infinite dimensional Hilbert space $X$.
Or is there any other proof of this fact?
You initial thoughts are technically correct but not the way to go to solve your questions. Rather, consider a sequence $\{x_n\}\subset U$ with $x_n\rightarrow p$ and $p\notin U$.
Then there is an $N$ with $x_N\neq p$ and $\|x_N-p\| < \frac{1}{3}$. Further there is a $K$ such that $\|x_K-p\| < \|x_N-p\| < 1/3.$ This implies $\|x_N-x_K\|< \frac{2}{3}$.
Finally show that this last inequality is not possible for elements of $U$ to get a contradiction.
