Let $S$ be a square formed by the four vertices $(1,1),(1.-1),(-1,1)$ and $(-1,-1)$. Let the region $R$ be the set of points inside $S$ which are closer to the center than any of the four sides. Find the area of the region $R$.
1. **Identify the region \( R \)**:
   The region \( R \) is defined as the set of points inside the square \( S \) that are closer to the center (origin) than to any of the four sides of the square. The vertices of the square \( S \) are \((1,1)\), \((1,-1)\), \((-1,1)\), and \((-1,-1)\).

2. **Distance from the center**:
   The distance from any point \((x, y)\) to the center (origin) is given by:
   \[
   d_{\text{center}} = \sqrt{x^2 + y^2}
   \]

3. **Distance from the sides**:
   The distances from the point \((x, y)\) to the sides of the square are:
   \[
   d_{\text{left}} = |x + 1|, \quad d_{\text{right}} = |x - 1|, \quad d_{\text{top}} = |y - 1|, \quad d_{\text{bottom}} = |y + 1|
   \]

4. **Set up inequalities**:
   For the point \((x, y)\) to be closer to the center than to any of the sides, the following inequalities must hold:
   \[
   \sqrt{x^2 + y^2} \leq |x - 1|
   \]
   \[
   \sqrt{x^2 + y^2} \leq |x + 1|
   \]
   \[
   \sqrt{x^2 + y^2} \leq |y - 1|
   \]
   \[
   \sqrt{x^2 + y^2} \leq |y + 1|
   \]

5. **Analyze each inequality**:
   - For \(\sqrt{x^2 + y^2} \leq |x - 1|\):
     \[
     x^2 + y^2 \leq (x - 1)^2
     \]
     \[
     x^2 + y^2 \leq x^2 - 2x + 1
     \]
     \[
     y^2 \leq -2x + 1
     \]
     \[
     y^2 + 2x \leq 1
     \]

   - For \(\sqrt{x^2 + y^2} \leq |x + 1|\):
     \[
     x^2 + y^2 \leq (x + 1)^2
     \]
     \[
     x^2 + y^2 \leq x^2 + 2x + 1
     \]
     \[
     y^2 \leq 2x + 1
     \]
     \[
     y^2 - 2x \leq 1
     \]

   - For \(\sqrt{x^2 + y^2} \leq |y - 1|\):
     \[
     x^2 + y^2 \leq (y - 1)^2
     \]
     \[
     x^2 + y^2 \leq y^2 - 2y + 1
     \]
     \[
     x^2 \leq -2y + 1
     \]
     \[
     x^2 + 2y \leq 1
     \]

   - For \(\sqrt{x^2 + y^2} \leq |y + 1|\):
     \[
     x^2 + y^2 \leq (y + 1)^2
     \]
     \[
     x^2 + y^2 \leq y^2 + 2y + 1
     \]
     \[
     x^2 \leq 2y + 1
     \]
     \[
     x^2 - 2y \leq 1
     \]

6. **Combine the inequalities**:
   The region \( R \) is bounded by the intersection of the regions defined by these inequalities:
   \[
   y^2 + 2x \leq 1
   \]
   \[
   y^2 - 2x \leq 1
   \]
   \[
   x^2 + 2y \leq 1
   \]
   \[
   x^2 - 2y \leq 1
   \]

7. **Symmetry and shape of the region**:
   The region \( R \) is symmetric about both the x-axis and y-axis. The boundaries of \( R \) are parabolas. The region \( R \) is a square with vertices at \((\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})\), \((\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}})\), \((- \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})\), and \((- \frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}})\).

8. **Calculate the area**:
   The side length of the square is the distance between \((\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})\) and \((\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}})\), which is:
   \[
   2 \times \frac{1}{\sqrt{2}} = \sqrt{2}
   \]
   The area of the square is:
   \[
   (\sqrt{2})^2 = 2
   \]

The final answer is \(\boxed{2}\).