1. Let $\sin^{-1}x = 2\sin^{-1}a.$ For this equation to have a real solution, both inverse sine expressions must be defined.

2. Domain conditions:

• $\sin^{-1}a$ is defined only when $-1 \le a \le 1.$
• Also, since $\sin^{-1}x$ always lies in the principal value range $\left[-\frac{\pi}{2},\frac{\pi}{2}\right],$ we must have $2\sin^{-1}a \in \left[-\frac{\pi}{2},\frac{\pi}{2}\right].$

3. Therefore, $-\frac{\pi}{2} \le 2\sin^{-1}a \le \frac{\pi}{2}.$ Divide throughout by $2$: $-\frac{\pi}{4} \le \sin^{-1}a \le \frac{\pi}{4}.$

4. Now apply sine on all sides. Since $\sin\theta$ is increasing on $\left[-\frac{\pi}{2},\frac{\pi}{2}\right],$ this gives $\sin\left(-\frac{\pi}{4}\right) \le a \le \sin\left(\frac{\pi}{4}\right).$ So, $-\frac{1}{\sqrt{2}} \le a \le \frac{1}{\sqrt{2}}.$ Hence, $|a| \le \frac{1}{\sqrt{2}}.$

5. Therefore the equation has a solution exactly when $\boxed{|a| \le \frac{1}{\sqrt{2}}}.$

6. Checking options:

• A: $|a| \ge \frac{1}{\sqrt2}$ — incorrect
• B: $\frac12 < |a| < \frac{1}{\sqrt2}$ — incomplete
• C: all real values of $a$ — incorrect
• D: $|a| \le \frac{1}{\sqrt2}$ — correct

So the correct option is D.