$$
a_{n+1}^2-a_n^2=6+a_{n}-a_n^2=(3-a_{n})(2+a_n)
$$
If $a_n>3, a_{n+1}>\sqrt{6+3}=3$. So by induction $a_n>3\;\forall\;n$ and
$a_{n+1}^2<a_{n}^2\;\forall\;n$. And only possible limit is the positive solution of $x^2=6+x$.
Here is a different approach that avoids induction by using Geometric Series and the Squeeze Theorem.
First, rewrite the square roots as exponents. The limit of the sequnce is given by:
$$\lim_{n\rightarrow\infty}c_{n+1}=\large2^{\sum\limits_{n=0}^{\infty}\frac{1}{2}(\frac{1}{2^n})}$$
We are only interested in whether or not the sequence converges, which is done by looking at the terms of the sequence where $n$ is large.
The only difference the sequence above and the one which you have defined is that the initial term $c_{1}=\sqrt{2}$ is greater than $1$. Notice that each $c_{n}$ is larger than each $a_n$. So if we have two sequences congerving to the same limit, and $a_{n}$ is between these two sequences for each $n\in\mathbb{N}$, then $(a_{n})_{n=1}^{\infty}$ will converge to the same limit. The tables below should explain this numerically in some sense.
\begin{array}{|c|c|}
\hline
n & 1 & 2 & 3 & 4 \\ \hline
b_{n} & 0.00000... &1.41421... & 1.74776... & 1.96157... \\ \hline
\end{array}
\begin{array}{|c|c|}
\hline
n & 1 & 2 & 3 & 4 \\ \hline
a_{n} & 1.000000... & 1.73205... & 1.93185... & 1.98289... \\ \hline
\end{array}
\begin{array}{|c|c|}
\hline
n & 1 & 2 & 3 & 4 \\ \hline
c_{n} & 1.41421... & 1.74776... & 1.96157... & 1.999037... \\ \hline
\end{array}
Furthermore, in hopes of using the Squeeze Theorem, we intend to "squeeze" $(a_{n})_{n=1}^{\infty}$ between two other sequences (as depicted in the tables) for each $n\in\mathbb{N}$. So now for the lower sequence, we just need to set $b_{1}=0$ and then define each $b_{n+1}$ the same way as we did for the other two sequences. This will give us
$b_{n}\leq a_{n}\leq c_{n}$ for all $n\in\mathbb{N}$.
Now, since the infinite series defined in the limit of $c_{n}$ can be written as:
$$\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\cdots}}}}=\sum\limits_{n=0}^{\infty}\frac{1}{2}\Bigl(\frac{1}{2}\Bigr)^{n}=\frac{1}{2}\Bigl(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots\Bigr)$$
Then we can use the formula for Geometric Series to see that this infinite series converges to $1$:
$$\sum\limits_{n=0}^{\infty}\frac{1}{2}\Bigl(\frac{1}{2}\Bigr)^{n}=\Bigl(\frac{1}{2}\Bigr)\dfrac{1}{1-(1/2)}=1$$
So $(c_{n})_{n=1}^{\infty}$ converges to $2$, and the same is true for $(b_{n})_{n=1}^{\infty}$ since each $b_{n}=c_{n-1}$ for all $n\geq 2$
Therefore by the Squeeze Theorem, the sequence $(a_{n})_{n=1}^{\infty}$ converges with $\lim\limits_{n\rightarrow\infty}a_{n+1}=\lim\limits_{n\rightarrow\infty}a_{n}=2$
Best Answer
Prove, by induction, that $a_n \geq 0$ for all $n$. In fact $a_n \geq 2$ for all $n$.