We are given the equation:
\[ |x + y| + |x - y| = 2 \]
Case 1: $x + y \ge 0$ and $x - y \ge 0$. This simplifies to:
\[ (x + y) + (x - y) = 2 \implies 2x = 2 \implies x = 1 \]
Substituting $x = 1$ into the conditions $x + y \ge 0$ and $x - y \ge 0$ yields:
\[ 1 + y \ge 0 \quad \text{and} \quad 1 - y \ge 0 \]
Solving these inequalities gives $y \ge -1$ and $y \le 1$. Thus, $y$ can be $-1, 0, 1$, providing 3 solutions when $x = 1$.
Case 2: $x + y \ge 0$ and $x - y \le 0$. This simplifies to:
\[ (x + y) + -(x - y) = 2 \implies (x + y) + (-x + y) = 2 \implies 2y = 2 \implies y = 1 \]
Substituting $y = 1$ into the conditions $x + y \ge 0$ and $x - y \le 0$ yields:
\[ x + 1 \ge 0 \quad \text{and} \quad x - 1 \le 0 \]
Solving these inequalities gives $x \ge -1$ and $x \le 1$. Thus, $x$ can be $-1, 0, 1$, providing 3 solutions when $y = 1$.
Case 3: $x + y \le 0$ and $x - y \ge 0$. This simplifies to:
\[ -(x + y) + (x - y) = 2 \implies (-x - y) + (x - y) = 2 \implies -2y = 2 \implies y = -1 \]
Substituting $y = -1$ into the conditions $x + y \le 0$ and $x - y \ge 0$ yields:
\[ x - 1 \le 0 \quad \text{and} \quad x + 1 \ge 0 \]
Solving these inequalities gives $x \le 1$ and $x \ge -1$. Thus, $x$ can be $-1, 0, 1$, providing 3 solutions when $y = -1$.
Case 4: $x + y \le 0$ and $x - y \le 0$. This simplifies to:
\[ -(x + y) + -(x - y) = 2 \implies (-x - y) + (-x + y) = 2 \implies -2x = 2 \implies x = -1 \]
Substituting $x = -1$ into the conditions $x + y \le 0$ and $x - y \le 0$ yields:
\[ -1 + y \le 0 \quad \text{and} \quad -1 - y \le 0 \]
Solving these inequalities gives $y \le 1$ and $y \ge -1$. Thus, $y$ can be $-1, 0, 1$, providing 3 solutions when $x = -1$.
Conclusion: Combining all four cases, we find a total of $3 + 3 + 3 + 3 = 12$ distinct integer solutions. The correct answer is 12.