Show $X=\left\{x \in [0,1]: x \neq \frac1n\text{ for any }n \in \Bbb N\right\}$ is neither compact nor connected I am stuck on the following question:  

Let $X=\{x \in [0,1]: x \neq \frac1n: n \in \Bbb N\}$ be given the subspace topology. Then I have to prove that $X$ is neither compact nor connected. 

Can someone give some explanation about how to tackle it? Thanks in advance for your time.
 A: Let's consider a picture of $X$:



For convenience / as a suggestion, define
$$U_n=(\tfrac{1}{n+1},\tfrac{1}{n})$$
so that
$$X=U_1\cup U_2\cup\cdots\cup\{0\}.$$
Can you find an infinite open cover of $X$ with no subcover (demonstrating that $X$ is not compact)?
Can you find two disjoint non-empty open sets of $X$ whose union is $X$? (recall that any union of open sets is itself an open set)
Edit: I wrote this answer mistaking thinking that $\{0\}$ is open in $X$, when of course it is not. Therefore, the answer that was (intentionally) heavily suggested by what I wrote for compactness is not actually correct. TonyK's suggestion below, to give an infinite open cover of one of the $U_n$'s and then cover the rest of $X$, is correct. For connectedness, you can still break $X$ into $U_1$ and $U_2\cup U_3\cup\cdots\cup\{0\}$ because the latter is just $X\cap (-\frac{1}{2},\frac{1}{2})$, and hence is indeed an open set of $X$.
A: Here is a quick way of seeing this:


*

*$A\subseteq\Bbb R$ is connected if and only if $A$ is an interval, but $X$ is not an interval since $\frac1\pi,\frac1e\in X$ but $\frac1\pi<\frac13<\frac1e$ and $\frac13\notin X$.

*If $A\subseteq\Bbb R$ is compact then it is closed, therefore whenever $(a,b)\subseteq A$ is an open interval we have that $[a,b]\subseteq A$ as well. But again, $(\frac13,\frac12)\subseteq X$, whereas $\frac13\notin X$ so $[\frac13,\frac12]\nsubseteq X$.
A: To disprove compactness, show that $X$ isn't closed in $\Bbb R$. Alternately, cover $X$ by an appropriate collection $\mathcal{U}$ of infinitely-many open real intervals with some fixed negative lower endpoint. The collection $\mathcal V:=\{X\cap U:U\in\mathcal U\}$ should be an open cover of $X$ without finite subcover, if you've chosen well.
To disprove connectedness, find a relatively open subset of $X$ that is also relatively closed. I recommend that you find a closed interval of $\Bbb R$ whose intersection with $X$ is an open interval in $\Bbb R$.
