1. **State the problem:** Solve the inequality $$\left(\frac{1}{2}\right)^{x - 2\sqrt{x} - 3} > 1.$$\n\n2. **Recall the properties of exponents:** For a base $0 < a < 1$, the function $a^t$ is decreasing in $t$. Also, $1 = a^0$ for any $a \neq 0$.\n\n3. **Rewrite the inequality:** Since the base $\frac{1}{2}$ is between 0 and 1, the inequality $$\left(\frac{1}{2}\right)^u > 1$$ is equivalent to $$u < 0,$$ where $$u = x - 2\sqrt{x} - 3.$$\n\n4. **Set up the inequality:** $$x - 2\sqrt{x} - 3 < 0.$$\n\n5. **Substitute:** Let $$t = \sqrt{x} \geq 0,$$ then $$x = t^2.$$ Substitute into the inequality:\n$$t^2 - 2t - 3 < 0.$$\n\n6. **Solve the quadratic inequality:**\nFactor the quadratic:\n$$t^2 - 2t - 3 = (t - 3)(t + 1).$$\n\n7. **Analyze the inequality:**\n$$(t - 3)(t + 1) < 0.$$\nThis product is negative when one factor is positive and the other is negative. Since $t \geq 0$, $t + 1 > 0$ always. So the inequality reduces to:\n$$t - 3 < 0 \implies t < 3.$$\n\n8. **Domain restriction:** Since $t = \sqrt{x} \geq 0$, the solution for $t$ is:\n$$0 \leq t < 3.$$\n\n9. **Back-substitute for $x$:**\n$$0 \leq \sqrt{x} < 3 \implies 0 \leq x < 9.$$\n\n10. **Final answer:**\n$$\boxed{0 \leq x < 9}.$$