Prove that $a_n \in [0,2)$ 
Let $(a_n)_{n \in \mathbb{N}}$ be a sequence, with $a_0=0$, $a_{n+1}=\frac{6+a_n}{6-a_n}$.
Prove that $a_n \in [0,2)$ $\forall n \in \mathbb{N}$

Here's what I did:
I tried to prove this by induction:
Base case:
$0 \leq a_0 (=0)  < 2$.
Inductive step:
Suppose that $0 \leq a_n  < 2$
So $$\begin {split} 0 \leq a_n  < 2 &\iff 0 \geq -a_n > -2 \\ &\iff 6 \geq 6-a_n > 6-2 \\ &\iff \frac{1}{6} \geq \frac{1}{6-a_n} > \frac{1}{4} \\ &\iff \frac{a_n}{6}+1 \geq \frac{6+a_n}{6-a_n} > \frac{3}{2} + \frac{a_n}{4} \end{split}$$ 
To be fair I have no idea if this is going somewhere.

First, let's rewrite the recurrence formula:
$$a_{n+1}=\frac{6+a_n}{6-a_n}=\frac{a_n-6+12}{6-a_n}=-1+\frac{12}{6-a_n}$$
Now, we have:
$$0\leq a_n<2$$
$$6\geq 6-a_n > 4$$
$$2\leq \frac{12}{6-a_n} < 3$$
$$1\leq -1+\frac{12}{6-a_n}=a_{n+1} < 2$$
Thus, $a_{n+1} \in [1, 2)$ and since $[1, 2) \subset [0, 2)$, $a_{n+1} \in [0, 2)$.
