1. Define the problem and notation:
We are given a $10 \times 10$ square on an infinite squared plane. Each unit segment forming these squares is colored, and the border of every square with sides on grid lines contains segments of at most two colors. We need to determine the maximum number of colors that can appear in this coloring.

2. Introduce the notation for segments:
- Let $R(i,j)$ be the segment that runs to the right from $(i,j)$, connecting $(i,j)$ with $(i+1,j)$.
- Let $U(i,j)$ be the segment that runs upwards from $(i,j)$, connecting $(i,j)$ with $(i,j+1)$.

3. Describe the segments on the board:
- The horizontal segments $R(i,j)$ exist for $0 \leq i \leq 9$ and $0 \leq j \leq 10$.
- The vertical segments $U(i,j)$ exist for $0 \leq i \leq 10$ and $0 \leq j \leq 9$.

4. Construct a coloring with 11 colors:
- Color all segments black (let's call this "Color 0").
- For $0 \leq i \leq 9$, color $R(i,i)$ with color $i+1$.
- This ensures that no square contains $R(i,i)$ and $R(k,k)$ for $i < k$ because these segments are horizontal and would require the square to have a height and width of $k-i$, which is not possible.

5. Prove that 11 is the maximum number of colors:
- Case 1: All horizontal segments are the same color (e.g., black):
- Two vertical segments at the same height, $U(i,z)$ and $U(j,z)$, cannot have two different colors that aren't black because there would be a square with edges along both segments and a black horizontal line.
- Thus, for every row $z$, there is at most one non-black color among the vertical segments at this height.
- Overall, this gives at most 11 colors (1 black + 10 rows).

- Case 2: Each row is monochromatic but different rows may have different colors:
- There can be at most two different row colors (proof omitted for brevity).
- All rows are either all black or all white.
- There must be two adjacent rows of different colors, say rows $z$ and $z+1$.
- The vertical segments $U(i,z)$ cannot have any color other than black/white, and the verticals at every other height $z'$ can have only one additional color.
- Thus, we cannot exceed 11 colors (black, white, and at most 9 new colors from each height other than $z$).

- Case 3: Rows and columns are not all monochromatic:
- Suppose there is a row with both black and white horizontal segments.
- There is an adjacent pair $L(i,j)$ and $R(i,j)$ where one is black and the other is white.
- Every vertical segment $U(x,y)$ where $x \neq i$ must be either black or white.
- All columns except for column $i$ are black/white.
- Since we have a non-monochromatic column, all rows must have colors $C/D$ for some colors $C$ and $D$, except for maybe one row.
- This enables at most 8 colors (black, white, $C$, $D$, two colors at column $i$, and two colors at the exempt row).

6. Conclusion:
- The maximum number of colors that can appear in this coloring is 11.

The final answer is $\boxed{11}$