1. **State the problem:** Solve the equation $$\sqrt{x^{2} - 8x + 16} + \sqrt{x^{2} - 4x + 4} = 2.$$\n\n2. **Recognize perfect squares:** Notice that \(x^{2} - 8x + 16 = (x-4)^2\) and \(x^{2} - 4x + 4 = (x-2)^2\). So the equation becomes $$\sqrt{(x-4)^2} + \sqrt{(x-2)^2} = 2.$$\n\n3. **Simplify square roots:** Since square root of a square is the absolute value, we have $$|x-4| + |x-2| = 2.$$\n\n4. **Analyze absolute values:** The expression depends on the position of \(x\) relative to 2 and 4. We consider intervals:\n- For \(x \leq 2\): \(|x-4| = 4 - x\), \(|x-2| = 2 - x\). Sum: \(4 - x + 2 - x = 6 - 2x\). Set equal to 2: \(6 - 2x = 2 \Rightarrow 2x = 4 \Rightarrow x = 2\).\n- For \(2 < x < 4\): \(|x-4| = 4 - x\), \(|x-2| = x - 2\). Sum: \(4 - x + x - 2 = 2\). This is always true for any \(x\) in \((2,4)\).\n- For \(x \geq 4\): \(|x-4| = x - 4\), \(|x-2| = x - 2\). Sum: \(x - 4 + x - 2 = 2x - 6\). Set equal to 2: \(2x - 6 = 2 \Rightarrow 2x = 8 \Rightarrow x = 4\).\n\n5. **Combine solutions:** From the intervals, solutions are \(x = 2\), all \(x\) in \((2,4)\), and \(x = 4\). So the solution set is \([2,4]\).\n\n6. **Check domain:** Since the original expressions are square roots, the radicands must be non-negative. Both \((x-4)^2\) and \((x-2)^2\) are always non-negative, so domain is all real numbers.\n\n7. **Final answer:** $$\boxed{[2,4]}.$$