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.