Prove $\bigcup\limits_{n \in \mathbb{ N} }{\left[0,1-\frac{1}{n}\right]} = [0,1)$ First, I apologize for my English, it is not my native language.
To solve this exercise I started by proving that 
$\displaystyle\bigcup_{n \in \mathbb{ N} }{\left[0,1-\frac{1}
 {n}\right]} \subseteq [0,1)$:
Let $x \displaystyle\bigcup_{n \in \mathbb{ N} }{\left[0,1-\frac{1}
 {n}\right]}$.
Therefore, there exist some $N \in \mathbb{N}$ such that
$x \in \left[0,1-\frac{1}{N}\right]$
Since $N \in \mathbb{N}$, $\frac{1}{N} > 0$. (Note that in my case the natural numbers start at 1)
Thus $0 \leq x \leq 1- \frac{1}{N} < 1$,
Then $x \in [0,1)$
This is, $\left[0,1-\frac{1}{N}\right] \subseteq [0,1)$
Then, $\displaystyle\bigcup_{n \in \mathbb{ N} }{\left[0,1-\frac{1}{n}\right]} \subseteq [0,1)$.
Edit
(I used the suggestions received to finish the second part of the proof. Is this well?)
To obtain the opposite inclusion, suppose $x\subseteq [0,1)$ and choose $ \varepsilon =1-x >0$.
By the Archimedean property follows that there exist $N \in \mathbb{N}$ such that
$\frac{1}{N}<1-x$
Thus, $0 \leqslant x<1-\frac{1}{N}$.
Therefore, $x \in \left[0,1-\frac{1}{N}\right]$. 
This means, $x \in \displaystyle\bigcup_{n \in \mathbb{ N} }{\left[0,1-\frac{1}{n}\right]}$.
We conclude that
$[0,1) \subseteq \displaystyle\bigcup_{n \in \mathbb{ N} }{\left[0,1-\frac{1}
 {n}\right]}$.
 A: Your first part of the proof is... OK. It's not perfect. It has some flaws where, sure, because the statement is simple, I know what you were trying to say, but you didn't really say that, and you didn't make your intention clear. For example:

I started by proving that 
$\displaystyle\bigcup_{n \in \mathbb{ N} }{\left[0,1-\frac{1}{n}\right]} \subseteq [0,1)$.
I tried it this way:
Let $x\in \left[0,1-\frac{1} {N}\right]$ for some $N \in \mathbb{N}$.

Why did you select this $x$? Usually, when we are proving $A\subseteq B$, we must take some $x\in A$, but you didn't do that. You selected $x$ from a subset of $A$.
Instead, the beginning of your proof should be something like this:

Let $$x\in \bigcup_{n\in\mathbb N}\left[0, 1-\frac1n\right].$$ Therefore, there exists some $N\in\mathbb N$ such that $$x\in\left[0, 1-\frac1N\right].$$

The rest of this part of the proof is then OK.

To prove the second part, you need to prove that if $x\in[0,1)$, then $$x\in \bigcup_{n\in\mathbb N}\left[0, 1-\frac1n\right].$$ To do that, think about what $x<1$ means, and try to find some $N$ such that $x<1-\frac{1}{N}$.
