Darwin takes an $11\times 11$ grid of lattice points and connects every pair of points that are 1 unit apart, creating a $10\times 10$ grid of unit squares. If he never retraced any segment, what is the total length of all segments that he drew?

[i]Ray Li.[/i]

[hide="Clarifications"][list=1][*]The problem asks for the total length of all *unit* segments (with two lattice points in the grid as endpoints) he drew.[/list][/hide]
1. **Understanding the Grid**:
   - The grid is an \(11 \times 11\) grid of lattice points.
   - Each pair of points that are 1 unit apart is connected, forming a \(10 \times 10\) grid of unit squares.

2. **Counting Horizontal Segments**:
   - Each row of the grid has 10 horizontal segments.
   - There are 11 rows in total.
   - Therefore, the total number of horizontal segments is:
     \[
     10 \text{ segments/row} \times 11 \text{ rows} = 110 \text{ horizontal segments}
     \]

3. **Counting Vertical Segments**:
   - Each column of the grid has 10 vertical segments.
   - There are 11 columns in total.
   - Therefore, the total number of vertical segments is:
     \[
     10 \text{ segments/column} \times 11 \text{ columns} = 110 \text{ vertical segments}
     \]

4. **Total Length of All Segments**:
   - Each segment is 1 unit long.
   - The total number of segments (both horizontal and vertical) is:
     \[
     110 \text{ horizontal segments} + 110 \text{ vertical segments} = 220 \text{ segments}
     \]
   - Since each segment is 1 unit long, the total length of all segments is:
     \[
     220 \text{ segments} \times 1 \text{ unit/segment} = 220 \text{ units}
     \]

\(\blacksquare\)

The final answer is \(\boxed{220}\)