Are $x$ and $y$ required to be integers? Because if they are not, then the answer cannot be $4$: take $x=6.5$ and $y=4$. Then $x+y = 10.5\lt 11$, but $x-y = 2.5 \lt 4$. In fact, taking $x=6+\epsilon$ and $y=5-(\epsilon+\delta)$ with $\epsilon\gt 0$ and $\delta\gt 0$ arbitrary but positive, and this gives you $x+y = 11-\delta\lt 11$, but $x-y = 1+2\epsilon+\delta$, so we can make $x-y$ as close to $1$ as we like (without it ever being equal to $1$); on the other hand, since $x\gt 6$ and $x+y\lt 11$, then $y\lt 11-x \lt 11-6 = 5$, so $x-y \gt 6-5 = 1$. Thus, the difference is always greater than $1$; but since we can make it as close to $1$ as we want, the best possible lower bound for $x-y$ is $1$ (without it actually being achieved).
However, if $x$ and $y$ are required to be integers, then the smallest value of $x-y$ will occur when $x$ is as small as possible and $y$ as large as it possibly can be given that.
Added: To see why, first suppose that $x+y\lt 11$ with $x\gt 6$, but $y$ is not as large as it can be; that is, that $x+(y+1)$ is also less than $11$. Setting $y'=y+1$, you have that $x-y' = x-y-1 \lt x-y$, so you get a smaller value with $y'$. That means that in order to find the smallest possible value of $x-y$ for a given $x$, you may assume $y$ is as large as it possibly can be. Once you have that, you can check that larger values of $x$ will give you larger differences, so you also want $x$ to be as small as it possibly can be.