It's trivial, but for a different reason: a convergent sequence in any metric space is bounded. Suppose $\{x_n\}$ isn't bounded. Then no open ball centered at $x$ contains all points of the sequence. So for each $k$, we can find a point $x_{n_k}$ such that $d(x,x_{n_k})>k$. Take $\varepsilon=1$. Then for any $N$, we have a positive integer $n_k$ such that $d(x,x_{n_k})>k>1$. Hence, $x_n$ does not converge to $x$, a contradiction.
In fact, any Cauchy sequence is bounded. It's a good exercise to prove that as well.
Here's a more formal (and shorter) proof:
Let $(x_{n_k})$ be a convergent subsequence of $(x_n)$, which we know exists, and let $L$ be its limit. Then, given any $\varepsilon > 0$, there is some $K \in \mathbb{N}$ such that for all $k \geq K$, $|x_{n_k} - L| < \varepsilon$ (this is the definition of the limit).
Now, for any $n \geq n_K$, we will show than $|x_n - L| < \varepsilon$, which will complete the proof. For this purpose, suppose that this is not true. Then there is some $n \geq n_K$ such that $|x_n - L| \geq \varepsilon > |x_{n_K} - L|$.
First, note that if $x_1 \leq L$ then we must have $x_i \leq L$ for all $i$: if not, then we have some $i$, such that $x_1 \leq L < x_i$, but then for all $m > i$, $x_m \geq x_i > L$, so $|x_m - L| \geq |x_i - L| > 0$, so in particular, $(x_{n_k})\not\to L$, a contradiction. Symmetrically, if $x_1 \geq L$, then $x_i \geq L$ for all $i$.
Now, we have a problem: we have either $x_n \geq L + \varepsilon > x_{n_K} \geq L$ or $x_n\leq L - \varepsilon < x_{n_K} \leq L$, but then monotonicity gives us either $x_m \geq L + \varepsilon$ or $x_m \leq L - \varepsilon$ for all $m \geq n$, in particular for all but finitely many $n_k$, so $|x_{n_k} - L| \geq \varepsilon$ for all but finitely many $n_k$, so $(x_{n_k})\not\to L$, a contradiction.
Thus, we must have $|x_n - L| <\varepsilon$ for all $n \geq n_K$, hence $(x_n)\to L$.
Best Answer
Picking up from where you left off in the comment: Suppose x_n (defined above) is monotonically increasing.
The upper bound case is easy to show! $$x_1=1<21$$ Suppose $x_k < 21$ for some $k \in \mathbb{N}$. Then $x_{k+1} = \sqrt{20 + x_{k}} \leq \sqrt{41} \lt 21$, hence we have shown $x_n \lt 21$ for all $n \in \mathbb{N}.$
By the monotone convergence theorem, there is a limit. Find it now:
Remember that $\lim s_{n+1} = \lim s_{n}$.
Then $\lim x_{n+1} = \lim x_n$ satisfies $ x = \sqrt{20 + x}$ for a grand result of $x=5$. ($x$ here is the limit of the $x_n$ sequence)
As a whole, the process here is to show the monotonic increasing-ness of the sequence, use induction to show an upper bound with some constant substitution, and then use a recursive algebra trick to understand the limit's actual form.
I hope this supports your exam prep.