Why is $\bigcup_{n\geq 1}[0,1-1/n] \neq [0,1]$? Sorry but aren't we taking limits $\lim_{m \to \infty} \cup_{n =1 }^{m}[0,1-1/n] = [0,1]$? Why is this supposed to be equal to $[0,1)$?
 A: Aren't we taking limits?
No, we aren't taking limits.

Why is $\cup_{n\geq 1}[0,1-1/n] \neq [0,1]$?
Note that 
\begin{align}
&\qquad x\in \bigcup_{n \geq1 }[0,1-1/n]\\\\
&\Longrightarrow x\in[0,1-1/n]\text{ for some }n\geq 1\quad\text{(by definition of $\cup_{n\geq 1}$, see $(*)$)}\\\\
&\Longrightarrow 0\leq x \leq 1-1/n\text{ for some }n\geq 1\quad\text{(by definition of closed interval)}\\\\
&\Longrightarrow 0\leq x<1\quad \text{(because $1/n>0$ for all $n\geq 1$)}
\end{align}
This shows that
$$\bigcup_{n \geq1 }[0,1-1/n]\subset [0,1)\tag{A}$$
and thus
$$\bigcup_{n \geq1 }[0,1-1/n]\neq [0,1]$$

Why is this supposed to be qual to $[0,1)$?
Note that
\begin{align}
&\qquad x\in [0,1)\\\\
&\Longrightarrow 0\leq x < 1\quad\text{(by definition of $[0,1)$)}\\\\
&\Longrightarrow 0\leq x\text{ and }0<1-x\\\\
&\Longrightarrow 0\leq x\text{ and }\frac{1}{n}<1-x\text{ for some $n\geq 1$}\quad \text{(because $1/n\overset{n\to\infty}{\longrightarrow}0$)}\\\\
&\Longrightarrow 0\leq x \leq 1-1/n\text{ for some }n\geq 1\\\\
&\Longrightarrow x\in[0,1-1/n]\text{ for some }n\geq 1\\\\
&\Longrightarrow x\in \bigcup_{n \geq1 }[0,1-1/n]\\\\
\end{align}
This shows that
$$[0,1)\subset \bigcup_{n \geq1 }[0,1-1/n]\tag{B}$$
From $\text{(A)}$ and $\text{(B)}$ (and from the definition of equality of sets),
$$\bigcup_{n \geq1 }[0,1-1/n]=[0,1)$$

I can't get this picture out of my mind.
It seems your problem comes from the meaning of union. In general,
$$\bigcup_{n \geq1 } A_n=\{x\mid x\in A_n\text{ for some }n\geq 1\}$$
and thus
$$\bigcup_{n \geq1 }[0,1-1/n]=\{x\mid x\in [1,1-1/n]\text{ for some }n\geq 1\}.\tag{$*$}$$
It is not true that
$$\bigcup_{n \geq1 }[0,1-1/n]=[0,\lim(1-1/n)].$$

How can $\cap_{n \geq 1} (-1/n, 1/n) = \{0\}$? Why am I allow to apply limits over there, but not here?
Note that
$$\big(\lim(-1/n), \lim(1/n)\big)=(0,0)=\varnothing$$
and
$$\{0\}=[0,0].$$
So, you aren't just taking limits. You are also passing from open intervals to a closed interval.
A: $1 \in [0,1]$ but $1 \not\in \bigcup_{n=1}^\infty [0,1-1/n]$, for if $1 \in \bigcup_{n=1}^\infty [0,1-1/n],$ there would exist $n \in \mathbb{Z}_+$ such that $1\in [0,1-1/n],$ but that's impossible since $1>1-1/n$ for all $n \in \mathbb{Z}_+.$ 
A: "Limits" talk about "closeness" and what happens in the "long term". Sets talk about just yes-or-no, in-or-out - they don't care about "close". That's why $[0,1)$ is a different set than $[0,1]$ - even though the first set gets arbitrarily close to $1$, it never actually hits it, and for sets that's all that matters.
In general, $\bigcup_{n = 1}^{\infty}A_n$ is defined to be a set $A$ so that for any $x$, $x \in A$ iff $x \in A_n$ for some value of $n$. Again, don't think about limits or geometry here - this definition has nothing whatsoever to do with sets of reals in particular, so it doesn't even make sense to be talking about limits at all.
So $\bigcup_{n = 1}^{\infty}[0, 1 - 1/n]$ has to be that set $A$ so that $x \in A$ iff $x \in [0, 1 - 1/n]$ for some value of $n$. But $1$ is never in $[0,1 - 1/n]$ - for any particular value of $n$, the interval stops a little early. So $1 \notin A$. But every non-negative $x$ strictly less than $1$ does eventually fall into one of these intervals; for example, $0.9$ shows up for the first time at $n = 10$. So $A$ is the set $[0,1)$.
A: A union of infinitely many sets is not treated like a limit in the same way a sum of infinitely many numbers might. They are defined differently. An infinite sum is (usually) defined as the limit of the partial sums:
$$ a_1 + a_2 + \ldots := \lim_{n \to \infty} (a_1 + a_2 + ... + a_n) $$
but an infinite union is defined without reference to limits, and in fact is very different from a limit:
$$ \bigcup_{i \in I} A_i := \{a\colon a \in A_i \text{ for some } i \in I\} $$
That is, the set of all elements $a$ that are members of at least one of the $A_i$ sets.
Using this definition, $\bigcup_{n\geq 1}[0,1-1/n]$ does not contain $1$ simply because $1$ is not an element of any of the constituent sets. However, it is true that the closure of the set is $[0,1]$ (hence containing $1$), which feels more in line with our intuitions about limits, but this is because the notion of closure is a topological one (just like limits are).
To put a long story short, $1$ is not included because the definition of an infinite union does not involve any notion of "limit" (i.e. does not employ any notion of sets "approaching" anything).
A: Not to confuse anyone, but a lot of the previous (very good) answers talk to much in absolutes: "you cant talk about limits of sets", "the definition is...". There is actually nothing wrong with defining that infinite union as a limit
Any countable union can perfectly well be defined as limit of finite unions in this case. Actually in any case where you work with subsets of some set $X$ (in contrast to just sets). In this case the space of subsets can be tough of as the space of fucntions from $X$ to $\{0,1\}$ (for each point you return whether it is in or out) which is the same as the product $\{0,1\}^X$. On the set $\{0,1\}$ we have a natural choice for a topology, namely the discrete topology, and by the way also a natural order, namely $0\leq 1$. This structure caries over to the product $\{0,1\}^X$ giving an order $\subseteq$ and a topology to subsets.
Now limits do have meaning and the limit of the finite unions as the OP uses is his question is a correct expression for the infinite union that he is asking about. (This uses the fact that the sequence of sets is monotone)
A: Probably inadvertently, you are asking two quite different things: one in the title and one in the body of the question. And in both cases you are right. For the question in the title, the comment by Cave Johnson is the clearest that comes straight to the point.
However, in the body you are asking about this: $$\lim_{m\rightarrow\infty}\bigcup\limits_{n=1}^{m} [0,~1-\frac{1}{n}]$$
First of all, being $[0,1-1/n]\subset [0,1-1/m]$, when $n>m$, $\cup_{n=1}^m[0,1-1/n]=[0,1-1/m]$. So you are asking about this limit of sequence of sets:
$$\lim_{n\rightarrow\infty}[0,~1-\frac{1}{n}]$$
You can read  for instance Aubin Frankowska - Set-valued Analysis for the definition of the limit of sequences of sets.
The limit of a sequence of sets is defined when its lower limit equals its upper limit. When the limit is defined it is equal to the lower (and upper) limit.
The lower (resp. upper) limit of a sequence of sets $K_n$ is the set of the limits (resp. cluster points) of all sequences of points $x_n \in K_n$. For $K_n=[0,1-1/n]$, the lower and upper limits coincide with $[0,1]$, as you correctly guessed.
A: This isn't a silly question. It's just that you've discovered a false belief.

Conjecture. Suppose $a : \mathbb{N} \rightarrow \mathbb{R}$ is an order-reversing sequence that's bounded below and $b : \mathbb{N} \rightarrow \mathbb{R}$ is an order-preserving sequence that's bounded above. Then:
  $$\bigcup_{i \in \mathbb{N}}[a_i,b_i] = \left[\lim_{i \in \mathbb{N}} a_i,\lim_{i \in \mathbb{N}} b_i \right]$$

Disproof. Assume toward a contradiction this were true. Then 
$$\bigcup_{i \in \mathbb{N}}\left[\frac{1}{i+1},1 \right] = \left[\lim_{i \in \mathbb{N}} \frac{1}{i+1},\lim_{i \in \mathbb{N}} 1 \right]$$
So $$\bigcup_{i \in \mathbb{N}}\left[\frac{1}{i+1},1 \right] = \left[0,1 \right]$$
So
$$[0,1] \subseteq \bigcup_{i \in \mathbb{N}}\left[\frac{1}{i+1},1 \right]$$
Hence $$0 \in \bigcup_{i \in \mathbb{N}}\left[\frac{1}{i+1},1 \right]$$
Hence $$\mathop{\exists}_{i \in \mathbb{N}}\left(\frac{1}{i+1} \leq 0 \leq 1\right)$$
Hence $$\mathop{\exists}_{i \in \mathbb{N}}\left(\frac{1}{i+1} \leq 0\right)$$
Hence $$\mathop{\exists}_{i \in \mathbb{N}} \bot$$
Hence $$\bot,$$ a contradiction.
