Yes. You can achieve $ 11$ colors, and no more. In general for an $ n \times n$ board the answer is $ n+1$. Fun problem - thanks for "Problem of the Day" for floating it up. Assume the board stretches from $ (0,0)$ to $ (10,10)$ on the Cartesian plane. It'll be convenient to refer to the segments like this: $ R(i,j)$ is the segment that runs to the right from $ (i,j)$, namely it connects $ (i,j)$ with $ (i+1,j)$, an similarly for $ L(i,j)$ ("left"), $ U(i,j)$ ("up") and $ D(i,j)$ ("down"). So the segments on the board can be described as $ R(i,j)$ for $ 0 \leq i \leq 9$, $ 0 \leq j \leq 10$ and $ U(i,j)$ for $ 0 \leq i \leq 10$, $ 0 \leq j \leq 9$. How to use 11 colors: Make all the segments black (let's call this "Color 0"), except that $ R(i,i)$ has color $ i+1$ for $ 0 \leq i \leq 9$. It is obvious that no square contains $ R(i,i)$ and $ R(k,k)$ for $ i < k$: these segments are both horizontal, so if a square contained both of them it would have to have horizontal edges along row $ i$ and row $ k$, so it's height must be $ k-i$. But then its width is $ k-i$ as well, and since it contains the point $ (i,i)$ it cannot contain $ (k+1,k)$ which is the rightmost end of $ R(k,k)$. So this is a legal coloring of the segments on the board. Now the proof that $ 11$ is the highest number possible: First, let's suppose that all the horizontal segments are the same color - say, black. Now two vertical segments at the same height, say $ U(i,z)$ and $ U(j,z)$, cannot have two different colors that aren't black, since there's a square with edges along both segments as well as a black horizontal line. Thus for every row $ z$ there's a maximum of one non-black color among the vertical segments at this height. Overall this gives $ 11$ colors at most. Next, let's suppose that each row (the horizontal segments with the same $ y$ coordinate) is monochromatic (single-colored), but different rows may have different colors. Note that there can be at most two different row colors (proof: exercise ). So all rows are either all black or all white. There must be two adjacent rows of different colors, say rows $ z$ and $ z+1$. Now 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 as before - so once again we cannot exceed $ 11$ colors (black, white, and at most 9 new colors from each height other than $ z$). Those arguments obviously apply to columns as well, so we can assume that the rows are not all monochromatic and neither are all columns. Suppose then that there's a row that contains both black and white horizontal segments. Then there's an adjacent pair $ L(i,j)$, $ R(i,j)$ of segments one of which is black and the other is white. This implies that every vertical segment $ U(x,y)$ where $ x \neq i$ must be either black or white, since for each such segment we can find a square that contains it and also contains $ L(i,j)$ and $ R(i,j)$. Thus all columns except for column $ i$ are black/white. By the same argument, 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. Overall this enables $ 8$ colors at most (black, white, $ C$, $ D$ two colors at column $ i$ and two colors at the exempt row). In fact much more can be said here but for our purposes this is enough: there cannot be more than $ 11$ colors.

Yes. just an outline: assume that there are two side-by-side segments with colors 0,1;call WLOG R(i,j) and R(i+1,j). it follows that the only segments which can have colors different from 0,1 are : R(k,j+1)&R(k,j-1) with 0=<k<10 R(h,j+2)&R(h,j-2) with 0=<h<4 ,6=<h<10 R(l,j+3)&R(l,j-3) with 0=<l<3 , 7=<l<10 R(m,j+4)&R(m,j-4) with 0=<m<2 , 8=<m<10 R(n,j+5)&R(n,j-5) with 0=<n<1 , 9=<n<10 you can do some little case work and get 11 as the most number of colors one can arrives. for an n*n grid one can do the same and got n+1 the maximum number of colors for coloring the grid.

A nice problem,it is better than this years third on Tuymaada(Combo),but not hard.Firs,an example for $11$.Color all segments with color $1$ expect the first vertical one in the first row,the second vertical in the second row...the tenth vertical in the tenth row. Now,the proof that it can't be more.Consider all lines expect the border lines(which build the border of $10*10$).Now,we know that we color them with at most two colors.Now,let all colors that are not those be special colors.Note that we can't have $1$ vertical and $1$ horizontal line which are colored with distinct special colors.Now,from this we can assume that all special colors are all horizontal or vertical,so now it is easy that if we can't have two segments which are colored with distinct special colors and are in the same row or column,so from this we are finished.

I think there is easy solution with induction.