Remember that every point in a discrete topology is an open set. In other words, every x has an immediate predecessor and successor. These two elements bound x and make x an open set.
If you think about it, it's hard to build a valuation group that is not discrete. Here's why. Suppose G is not discrete at x. This means x is a cluster point of G. Consider any other point y and add y-x to all the points in G. This transformation preserves order, so y is also a cluster point. Since y was arbitrary, every point in G is a cluster point.
There are ordered groups that look like this, but they usually aren't valuation groups. Consider the group generated by 1 and sqrt(2). This is isomorphic to Z*Z, with the ordering of the reals. The group is countable, so it doesn't cover all the reals. If a/b is a rational number that is within ε of the square root of 2, then a-b×sqrt(2) is within ε of 0. This makes 0 a cluster point, and every point is a cluster point. In fact, G is dense in the real line. Clearly G does not have the discrete topology, but G isn't a valuation group either.
If R is a valuation ring and a pid, it produces a valuation group equal to Z. Every element of G has an immediate predecessor and successor, G is discrete, and R is a dvr.