5. The solution set of the inequality $x^{2}+\sqrt{2-x} \leqslant 2$ is
5. $\left[\frac{1-\sqrt{5}}{2}, 1\right]$.

The original inequality is
$$
\sqrt{2-x} \leqslant 2-x^{2} .
$$

By definition, $-\sqrt{2} \leqslant x \leqslant \sqrt{2}$, and $y=2-x^{2}$ and $y=\sqrt{2-x}$ are inverse functions of each other within $[-\sqrt{2}, \sqrt{2}]$. Draw the function graphs (as shown in Figure 2).
Figure 2
Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)\left(x_{1}<x_{2}\right)$ be the two different intersection points of the two curves within $[-\sqrt{2}, \sqrt{2}]$.
When $x_{1} \leqslant x \leqslant x_{2}$, we have $2-x^{2} \geqslant \sqrt{2-x}$. From Figure 2, it is easy to see that $-1<x_{1}<0,0<x_{2}<\sqrt{2}$.
Consider $2-x=\left(2-x^{2}\right)^{2}$, clearly, $x_{2}=1$ is a root. Removing $x-1$, we get $x^{3}+x^{2}-3 x+2=0$, and it is easy to see that $x=-2$ is another root. Since $x_{1} \neq-2$, we can remove $x+2$, resulting in $x^{2}-x-1=0$. Given $-1<x_{1}<0$, we know $x_{1}=\frac{1-\sqrt{5}}{2}$.
Therefore, the solution set of the original inequality is $\left[\frac{1-\sqrt{5}}{2}, 1\right]$.