Solution verification: Prove by induction that $a_1 = \sqrt{2} , a_{n+1} = \sqrt{2 + a_n} $ is increasing and bounded by $2$ I have the following recursive relation (sequence):
\begin{align}
a_1 = \sqrt{2}, \quad a_{n+1} = \sqrt{2 + a_n}
\end{align}
My Try:
I'm a little skeptical of my manipulations near the end but it looks like it works out. 
Base Case:
Let $n=1$ then
\begin{align}
&a_2 = \sqrt{2 + a_1} \\
&a_2 < 2 \\
&\sqrt{2 + \sqrt{2}} < 2
\end{align}
The base case holds. 
Induction hypothesis: 
Let $n=k$
$$a_1 = \sqrt{2} \quad a_{k+1} = \sqrt{2 + a_k} \quad a_{k+1} = \sqrt{2}$$
Induction Step: 
Now we have to prove that $a_{k+2} < 2$. Let $n=k+1$.
\begin{align}
a_{k+2} &= \sqrt{2 + a_{k+1}} \\
\implies a_{k+2} &= \sqrt{2 + \sqrt{2 + a_k}} \\
\end{align}
Now we have to show that $a_{k+2} < 2$. 
\begin{align}
a_{k+2} &< 2\\
\sqrt{2 + \sqrt{2 + a_k}} &< 2\\
2 + \sqrt{2 + a_k} &< 4 \\
\sqrt{2 + a_k} &< 2 \\
\end{align}
Q.E.D 
Are my steps correct? 
Thanks for your time!
 A: The title mentions proving that $(a_n)$ is increasing, but you don't seem to handle that. 
More importantly, although you understand that the key to proving bounded-ness is $\sqrt{2+\sqrt2}<2$, you really need to work on the "style" of your induction proof. See for instance my answer here : Proof by Induction: 
By writing the induction steps in full length and correctly, you should immediately see what's ok or wrong in your proof.
Can I suggest that you should at least really revise the part where you write "Let $n = k$". What does it mean? Both $n$ and $k$ are silent variables... 
What you want to say is rather: let $n$ be an integer $\ge 2$ and let's assume that $P_n$ is true. Then blah, blah, which shows that $P_n \implies P_{n+1}$.
PS: for any induction proof, I can't recommend enough to write down in full length what the property $P_n$ is and, why not, the domain of $n$. Here that would be (formatting being a matter of taste):
$$ (n \ge 2)\quad P_n \; : \; 0 < a_n < 2 \quad\text{and}\quad a_n > a_{n-1}$$
That should make checking your initial case and your general case more "mechanical" (and thus easier) - it also makes your TA/exam corrector's life easier.
Late edit: I have tried to provide some basic help on induction proof writing here : Proof writing: how to write a clear induction proof?
A: I think you're making a bit of confusion. On one hand you are never proving that the sequence is increasing, on the other hand your argument is a bit too complicated. Here is how I would go about proving it:


*

*$\mathbf{n = 1}$: First note that $a_1 < a_2$ if and only if
$$
a_1^2 = 2 < 2 + \sqrt{2} = a_2^2
$$
which is clearly true. Further, $a_2 < 2$ because $2 + \sqrt{2} < 4$.

*Assume that $a_{n-1} < a_n < 2$. Then we need to prove that $a_n < a_{n+1} < 2$. For the first part note that $a_n < a_{n+1}$ is equivalent to $a_n^2 < 2 + a_n$, i.e.
$$
a_n (a_n - 1) < 2
$$
which is true because $a_n < 2$ implies $a_n - 1 < 1$. For the second part, note that $a_{n+1} < 2$ is equivalent to
$$
2 + a_n < 4
$$
which is true because $a_n < 2$ by hypothesis.
A: Your base case and inductive step are both overly complicated. 
Base case: We have $a_1 = \sqrt{2}$, which is less than $2$.
Inductive step: Assume we have $a_k < 2$
Now $a_{k+1} = \sqrt{2 + a_k} < \sqrt{2 + \sqrt{2}} < \sqrt{2 + 2} = 2$.
It may also be useful to remark that the sequence is strictly positive, so it is also bounded below.

The error you are making is that your base case proves $a_2$ but it should prove $a_1$ and your inductive step proves $a_{k+2}$ when you should be proving $a_{k+1}$.
Also, I'm a bit confused by your inductive step. There's no need to write the definition of the sequence in the inductive step and I'm confused why you have $a_{k+1} = \sqrt{2}$. Did you mean $a_{k+1} < \sqrt{2}$? Because if so, your inductive step (if you are assuming it for $n = k$) should say $a_k < \sqrt{2}$. Indeed, the statement you are trying to prove is $\{P_k : \text{each } a_k \text{ is bounded by } 2\}$. So in the inductive step you assume $P_n$ which is "$a_n$ is bounded by $2$"
A: $b_i:= a_i^2$ Then $$ b_1=2,\ b_{n+1}=2+\sqrt{b_{n}}$$
Show that $b_i$ is an increasing sequence bounded by $4$ :
(1) $b_1< 4$ If $b_n<4$ then $b_{n+1}=2+\sqrt{b_n}< 2+2=4$.
(2) $b_2=2+\sqrt{2} > b_1$ If $b_n>b_{n-1}$ then $$
b_{n+1}=2+\sqrt{b_n} > 2+\sqrt{b_{n-1}}=b_n $$
