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.