Is the set $T = \{1,\frac12,\frac13,\ldots,\frac1n,\ldots\}$ closed? 
Exercise:Is the set $(T = \{1,\frac12,\frac13,\ldots,\frac1n,\ldots\})$ closed in $\mathbb{R}$?

I tried to answer this question by computing $\mathbb{R}\setminus T=\bigcup_\limits{2}^{\infty}(\frac{1}{n-1},\frac{1}{n})\cup(1,\infty)\cup(-\infty,1)$
So $\mathbb{R}\setminus T$ is the union of open sets hence open which implies $T$ is closed.
The problem arose when I was told that $T$ is not closed.
Question:
What am I doing wrong? How should I solve the question?
Thanks in advance!

Your answer does not work because $\mathbb{R}\setminus T$ does not contain the interval $(-\infty, 1)$. For example, $\frac12\in T$.
One way to solve your question: can you show that if $T$ is closed, you must have $0\in T$? 
