Let $T$ be the discrete topology on $\mathbb{R}$ (so that every set is open) and let $m$ be a measure such that $m(A) = 0$ if $A$ is countable and $m(A) = \infty$ if $A$ is uncountable, for $A \subseteq \mathbb{R}$.
Claim: this is actually a measure. There are two properties to check. First, the empty set is countable, and thus has measure zero, as required. Second, suppose that $(A_n : n \in \mathbb{N})$ is a sequence of disjoint subsets of $\mathbb{R}$. If all the sets are countable, then so is their union, and so $m(\bigcup_{n \in \mathbb{N}} A_n) = \sum_{n \in \mathbb{N}} m(A_n) = 0$ as desired. On the other hand, if at least one of the sets $A_n$ is uncountable, then so is their union, and so $m(\bigcup_{n \in \mathbb{N}} A_n) = \sum_{n \in \mathbb{N}} m(A_n) = \infty$ (recall $\infty +\infty = \infty$). This proves the claim.
Now we look at the the support of $m$. Every point has a countable neighborhood; in fact $\{x\}$ is a neighborhood of a point $x$. These neighborhoods all have measure zero, so the support of $m$ is empty. On the other hand the measure is not identically zero.
This sort of thing cannot happen on the usual topology of $\mathbb{R}$ because it is strongly Lindelöf. With the usual topology, the complement of the support, which naturally has a covering by open sets of measure zero, must have a covering by a sequence of open sets of measure zero, and thus the complement of the support itself has measure zero, as does all of its subsets.
The Wikipedia article suggests different sufficient conditions for the measure to ensure that the support has measure zero: that the space is Hausdorff and the measure is Radon (locally finite and inner regular). The pathological measure I constructed above is locally finite and the space is Hausdorff, but the measure is not inner regular.