Let's use mathematical induction to prove this.
The inequality clearly holds for base case $n = 0$: $ u_0 \le 0 + \frac{u_0}{2^0} $
Assume the inequality holds for $n - 1$: $ u_{n-1} \le n - 1 + \frac{u_0}{2^{n-1}} \tag{1} $
We have:
\begin{align*} u_n &= \sqrt{n + u_{n-1}} \\ \Rightarrow u_n^2 &= n + u_{n-1} \end{align*}
Using (1): $ u_n^2 \le n + n - 1 + \frac{u_0}{2^{n-1}} $
Rearrange to get: $ \frac{u_n^2 + 1}{2} \le n + \frac{u_0}{2^n} $
In a previous question of yours, you've seen that: $ a \le \frac{a^2 + 1}{2} $
Therefore: $ u_n \le \frac{u_n^2 + 1}{2} \le n + \frac{u_0}{2^n} $
Which is what we want to prove.