Four. (20 points) Let $S$ be the set of real numbers $x$ that satisfy the inequality $x^{2}+$ $\sqrt{2-x} \leqslant 2$. Find $S$.

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

By definition, we know
$-\sqrt{2} \leqslant x \leqslant \sqrt{2}$,
and $y=2-x^{2}$
is the inverse function of $y=\sqrt{2-x}$ within $[-\sqrt{2}, \sqrt{2}]$. Draw the function graphs, as shown in Figure 9.

Let $A\left(x_{1}, y_{1}\right)$ and $B\left(x_{2}, y_{2}\right)$ be the two distinct 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 9, it is easy to see that
$$
-1<x_{1}<0,0<x_{2}<\sqrt{2} \text {. }
$$

Consider $2-x=\left(2-x^{2}\right)^{2}$, clearly, $x_{2}=1$ is a root. Dividing by $x-1$ gives
$$
x^{3}+x^{2}-3 x-2=0 \text {. }
$$

It is easy to see that $x=-2$ is another root.
Since $x_{1} \neq-2$, we can divide by $x+2$ to get
$$
x^{2}-x-1=0 \text {. }
$$

Given $-1<x_{1}<0$, we have $x_{1}=\frac{1-\sqrt{5}}{2}$.
Thus, $S=\left\{x \left\lvert\, \frac{1-\sqrt{5}}{2} \leqslant x \leqslant 1\right.\right\}$ is the solution.