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 $[0,1)$?

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.
