Instead of just proving the boundedness, let me show you the logic to construct the proof.
Step 1 guess the answer.
In this step, you don't need to be rigorous. You just use whatever intuition to guide you for a potential candidate of answer.
Let's say you want an upper bound $L$.
What does it mean? It means you want an $L$ such that for any $x \in (0,L]$, $\sqrt{\alpha+x} \le L$.
Since $\sqrt{\alpha+x}$ is an increasing function in $x$, you want
$$\begin{align}&\sqrt{\alpha+L} \le L\\
\iff &\alpha + L \le L^2\\
\iff &\alpha \le L^2 - L\\
\iff &\alpha + \frac14 \le (L - \frac12)^2\\
\implies &\sqrt{\alpha + \frac14} + \frac12 \le L\tag{*}
\end{align}$$
Notice $\sqrt{\alpha + \frac14}$ is bounded above by $\sqrt{\alpha} + \sqrt{\frac14} = \sqrt{\alpha} + \frac12$. This means if you have chosen $L$ such that $\sqrt{\alpha} + 1 \le L$, $(*)$ will be satisfied.
Step 2 verify you answer.
Once you have a potential candidate for an answer, you need to verify
it. If your logic of finding the candidate is reasonable. You can
usually reverse the steps to prove it is indeed what you want.
Let $L = \sqrt{\alpha}+1$, for any $0 \le x \le L$, we have:
$$\sqrt{\alpha + x} \le \sqrt{\alpha + L} = \sqrt{\alpha+ \sqrt{\alpha}+1} < \sqrt{\alpha + 2\sqrt{\alpha}+1} = \sqrt{\alpha} + 1 = L$$
This means if you start with a $x_n \le L$, you will get $x_{n+1} = \sqrt{\alpha + x_{n}} \le L$.
Since $x_0 \le L$, the boundedness of all $x_n$ is proved.
BTW, the rest of your steps look fine.
First, let me expand on Pedro's answer. Suppose $na_n$ converges to 0. We claim that the sequence $\dfrac{1}{n+1}(a_1+2a_2+...+na_n)$ converges to 0. Let $\epsilon > 0$. Then there is $N$ so that for all $n>N, |na_n| < \epsilon$. Let $M = |a_1+2a_2+...+Na_N|$. Then for any $n > N$, $|a_1+2a_2+...+na_n| \leq M + |(N+1)a_{N+1}+...+na_n| < M+(n-N)\epsilon$. Therefore, whenever $n>N$, $\dfrac{1}{n+1}|a_1+2a_2+...+na_n|<\dfrac{M}{n+1}+(\dfrac{n}{n+1}-\dfrac{N}{n+1})\epsilon$. Let $N'$ be such that for all $n>N'$, $\dfrac{M}{n+1} < \dfrac{\epsilon}{2}$, so that $\dfrac{1}{n+1}|a_1+2a_2+...+na_n|<1.5\epsilon$.
So $x_n-s_n$ converges to 0.
Let $s$ be the limit of the sequence $(a_n)$.
Then $|x_n-s|=|x_n-s_n+s_n-s|\leq |x_n-s_n|+|s_n-s|$. Let $\epsilon > 0$ be given. Then for sufficiently large $n$ the term on the right is less than $\epsilon$, which shows that the sequence $(x_n)$ converges to $s$.
Best Answer
Here is a formal way considering $f(x) = \left( \sqrt{2}\right)^x$ and using MVP:
It follows $$0 \leq 2 - x_{n+1} < \left(\frac{4}{5}\right)^n (2-x_1)\stackrel{n \to \infty}{\longrightarrow} 0 \Rightarrow \boxed{\lim_{n \to \infty}x_n = 2}$$
To sum this up: