Explanation on induction proof of a sequence I don't understand two key steps of a proof that the following sequence is bounded and monotone:
$$(a_n) \text{ defined recursively} \,, \quad a_1 = 0 \,,  \quad a_{n+1} = \frac{a_n +2}{2}  \,. $$
Can anyone explain either of these two steps in the proof? Thank you very much for any help. Here is an image of the original exercise
1. First the proof shows that the sequence $(a_n)$ is bounded:

Induction basis: $a_0 = 0 < 2$
Induction hypothesis: $a_n < 2,n ∈ N$
Induction step:  $a_{n+1} < 2$
$$a_{n+1} = 0.5a_n+1 <  \frac{2}{2}+1 = 2$$

First question: by the induction hypothesis obviously the 2 was inserted as $a_n$. Why is that valid?
2. Then the proof shows that $(a_n)$ is monotone:

$$a_{n+1} - a_n = \frac{a_n+2}{2}-a_n=-\frac{a_n}{2}+1$$
From $a_n < 2$ it follows that $-1 < \frac{-a_n}{2}$.
Hence, $a_{n+1} - a_n > -1 + 1 = 0$.

Second question: why does the inequality in the last line hold?
 A: Basically the proof is somewhat confusing to follow because $a_n$ and related symbols can have slightly different (albeit related) meanings depending on the context. Sometimes when I get confused by an argument like that, I introduce ancillary symbols/variables to make clearer to myself which parts of the argument I need to pay most attention to and how they are related. This is what I have done when attempting to answer your questions below.
Regarding the first question:
You are right, it would be incorrect to assume that $a_n = 2$, that $a_n$ exactly equals $2$, because the induction hypothesis only says that $a_n < 2$, i.e. that $a_n$ is some number less than $2$.
So for any given/fixed $n$, define $x:= a_n$, by the induction hypothesis $x < 2$, and by definition of the sequence $(a_n)$, we have that $a_{n+1} = \frac{1}{2}(x+2)$. Because $x < 2$, it follows that $\frac{1}{2}(x + 2) < \frac{1}{2}(2+2) = 2$.
Does it makes sense explained that way?
Regarding the second question:
From the first line, they are saying that $a_{n+1} - a_n = y + 1$, where $y = -\frac{a_n}{2}$. I assume you understood the argument why $y > -1$.
Then because $a_{n+1} - a_n = y +1$, and $y > -1$, we have that $a_{n+1} - a_n > (-1) + 1 = 0$.
