Weird thing about z-transform and difference here is my doubt: we were told that the ROC of the Z-transform of the sum of two sequences is the intersection of the respective ROCs as the two of them are limited only if both of them are. Now I had to solve an exercise where I had to compute the z-transform of the difference of two sequences and establish the ROC. It looked like this: $u[n] - u[n-k]$ where $u[n]$ is the heaviside step function and k=10. I tried in two different ways: first one was by using the linearity of z-transform and I got that
$X(z) = \frac{z}{z-1} - z^{-10}\frac{z}{z-1} = \frac{z}{z-1} + z^{-10}\frac{z}{1-z}$ 
And it would look like we need to have for the first to be finite that $ \lvert z\rvert > 1$ and for the second one as well. Instead if I compute this by definition I get that $X(z) = \sum\limits_{k=0}^{9} z^{-k}$ that looks to be finite for each $z\neq 0$.
Why does it happen? Is there some particular configuration that caused this?

You are right. The ROC of the sum of $u[n]$ and $-u[n-10]$ is the full plane except $z=0$. And the ROC of each summand is $|z|>1$.

we were told that the ROC of the Z-transform of the sum of two sequences is the intersection of the respective ROCs

Not quite. More precisely: given the ROCs of two signals, the ROC of the sum is "at least" the intersection of the two. That is the best you can say, if you are not given more data. But it can be larger. In other words: $ROC(x_1 +x_2)\subseteq ROC(x_1) \cap ROC(x_2) $
In terms of zeros and poles (if we are dealing with rational Z-transforms - as it's usual, and as it's the case case), if our sequences are right-sized (zero for $n<0$) the ROC is given by $|z|> |z_0|$  where $z_0$ is the out-most pole.
Hence, the ROC of the sum will depend on the largest pole of the sum (intersection of ROCs)... unless there is a pole cancellation. That's what happen here: the pole at $z=1$ disappears.
