To argue that a point is not an accumulation point of a given set I want to show that $\mathbb Z^2$ has no accumulation points in $\mathbb R^2\backslash\mathbb Z^2$. Is this argument correct? In particular, have I correctly invoked the density property of $\mathbb R$?

For all $v\in\mathbb R^2\backslash\mathbb Z^2$, there is some
  $\epsilon >0$ (since $\mathbb R ^+$  is dense in itself) such
  that the (punctured) $\epsilon$-ball about $v$ does not intersect
  $\mathbb Z^2$.
(In other words: Because $\mathbb R^+$ is dense in itself, every
  point in $\mathbb R^2\backslash\mathbb Z^2$ has some 
  neighbourhood containing no point of $\mathbb Z^2$.)

 A: I don't see why you want to invoke a density argument. What you want to show is, given a point $p=(x,y)\in \mathbb R^2 - \mathbb Z^2$, that $p$ is not an accumulation point of $\mathbb Z^2$. So, you need to demonstrate that some $\epsilon >0$ exists such that $B_\epsilon(p)$ does not intersect $\mathbb Z^2$. Now, the points of $\mathbb Z^2$ define a grid in $\mathbb R^2$ and the point $p$ is inside some square determined by this grid. Let $\epsilon$ be the smallest of the distance from $p$ to each of the four corners of that square and now you can prove that this $\epsilon $ works. (You might want to draw some pictures if the argument is not clear.)
A: Your argument does not work. Meaning, the first line is what you want to prove, but you haven't proven it. (in general since being dense in itself is a trivial property that every space has, it will never be useful). 
Also note that just because there is a point from $\mathbb{R}^2$ in the neighborhood doesn't mean that there $\textit{isn't}$ a point from $\mathbb{Z}^2$ in the neighborhood as well (which is what you want to show). Finally it is worthwhile to note that you don't actually use anything about $\mathbb{Z}^2$ this is a dead giveaway that you are not finished. I mean think about it. If you are trying to show $A$ has no accumulation points in $B\setminus A$ and you don't use anything about $A$ then it is true for any $A$, (this is a point many students miss when first starting out, ie that their theorem is true for everything that has the properties they use in the proof NOT just what they state the theorem for.)
Give $p=(a, b)\in\mathbb{R}^2\setminus\mathbb{Z}^2$ we know that $a, b\in\mathbb{R}\setminus\mathbb{Z}$ so the idea is to choose $\epsilon$ so that $(a-\epsilon, a+\epsilon)\subset\mathbb{R}\setminus\mathbb{Z}$ as well, and similarly for $b$. 
