2. Let $a \in \mathbf{R}, A=\{x|x \in \mathbf{R}| x-a \mid, \leqslant 1\}$, $B=\left\{x|x \in \mathbf{R}| x-1 \mid, \leqslant a^{2}\right\}$. If $A$ is not a proper subset of $B$, then the range of values for $a$ is ( ).
(A) $-1 \leqslant a \leqslant 1$
(B) $a \leqslant-2$ or $a>1$
(C) $-2<a \leqslant 1$
(D) $-2 \leqslant a \leqslant 0$
2. C.

Obviously, $A=\{x \mid x \in \mathbf{R}, a-1 \leqslant x \leqslant a+1\}$,
$$
B=\left\{x \mid x \in \mathbf{R}, 1-a^{2} \leqslant x \leqslant a^{2}+1\right\},
$$
both $A$ and $B$ are non-empty.
Assuming $A$ is a proper subset of $B$, then we have
$$
\left\{\begin{array} { l } 
{ a ^ { 2 } + 1 > a + 1 , } \\
{ 1 - a ^ { 2 } \leqslant a - 1 }
\end{array} \text { or } \left\{\begin{array}{l}
a^{2}+1 \geqslant a+1, \\
1-a^{2}0, \\ (a-1)(a+2) \geqslant 0\end{array}\right.\text{ or } \left\{\begin{array}{l}a(a-1) \geqslant 0, \\ (a-1)(a+2)>0 .\end{array}\right.$
Solving, we get $\left\{\begin{array}{l}a>1 \text { or } a1 .\end{array}\right.$ Therefore, $a>1$ or $a \leqslant-2$.
However, the condition in the original problem is “$A$ is not a proper subset of $B$”, so we have $-2<a \leqslant 1$.