Give an example of a topological space $X$, and a connected component which is not open in $X$ 
Give an example of a topological space $X$, and a connected component which is not open in $X$

I know of the following theorems:


*

*Each connected component is closed;

*If a topological space has a finite number of connected components, then those components are open in $X$.


Because of the second theorem, I need to be on the lookout for a topological space with a infinite number of connected components. I was thinking of something like this:
$$X = \bigsqcup_{n\in \mathbb{N}} \left[n, \dfrac{3}{2} n\right[$$
But this doesn't seem to work since $\sqcup_{n\geqslant 2} [n, \frac{3}{2}[$ as an union of infinite closed sets seems to be closed again. 
The open/closed concept in a relative topology can be mindbending. Could someone help me to find an simple example?

The space $\displaystyle X = \Big \{\frac{1}{2}, \frac{1}{3}, \ldots, 0  \Big\}$ is completely disconnected. That is to say the connected components are the singletons $\displaystyle \Big \{\frac{1}{2}\Big\},  \Big \{\frac{1}{3}\Big\} \ldots $ all of which are open, and $\{0\}$ which is not. 
To show $\{0\}$ is not open observe the series $\frac{1}{2}, \frac{1}{3}, \ldots$ converges to zero, so is eventually in any open set about $0$. Therefore $\{0\}$ fails to be open. To show $\{0\}$ is a component observe that for each $n$ we can write $X=\Big \{\frac{1}{2}, \frac{1}{3}, \ldots, \frac{1}{n} \Big \} \cup \Big\{\frac{1}{n+1}, \frac{1}{n+2}, \ldots ,  0  \Big\}$. This demonstrates that $\displaystyle\frac{1}{n}$ and $0$ are in different components.
