Support of a measure on Borel sets not well defined. Let $(X, \tau$) be a topological space and let $B$ be the Borel $\sigma$-algebra generated by $\tau$. Let  $(X,B,\mu)$ be a measure space. Then the support of the distribution $\mu$ is defined as :
$$\operatorname{supp}({\mu})=\{x\in X : \mu(U)>0\text{ for all U open neighborhood of }x\}.$$ Now in the wikipedia article about  Radon measures, it is written

A common problem is to find a good notion of a measure on a
topological space that is compatible with the topology in some sense.
One way to do this is to define a measure on the Borel sets of the
topological space. In general there are several problems with this:
for example, such a measure may not have a well defined support.
Another approach to measure theory is to restrict to locally compact
Hausdorff spaces[...]

But I don't see why the support is not well defined when we use the above expression.
 A: Support  is defined in that article as the smallest closed set $S$ such that $\mu (S^{c})=0$. This is the usual definition of supprt and it is the most convenient. With this definition the complement of the support is the largest open set of measure $0$. If you consider the collection  of all open sets of measure $0$ their union is cetainly open but it may not have measure $0$ since an uncountable union of sets of measure $0$ need not have measure $0$. That is the reason why not every Radon measure has a support.  On any second countable topological space every Radon measure has  a support.
A: If the topological space is not second countable, the definition
$$\operatorname{supp}({\mu})=\{x\in X : \mu(U)>0\text{ for all $U$ open neighborhood of }x\}$$
can still be applied, but it may not yeld what is typically expected of the support of measure.
So while the definition of support of a measure as you posted, or as we can find here, (they are the same definition with slightly different wording) is a definition that can be applied to the general case, it is only useful (or meaningful) if there are some condition on the topology (such as being second countable).
Let us see a concrete simple example:
Let $X=[0,1]$ and $\tau = \{ A \subseteq [0,1] \}=2^X$. Please, note that $(X, \tau)$ is not second countable.
Let $\mathcal{B}$ be the Borel $\sigma$-algebra corresponding to $\tau$. So  $\mathcal{B}= 2^X$.
Let $\mu$ be a measure defined by, for any $E \in \mathcal{B}$, $\mu(E)=0$ if $E$ is countable and  $\mu(E)=+\infty$ if $E$ is uncountable. It is easy to check that $\mu$ is a measure.
Now what is the support of $\mu$?  Let us simply apply the definition:
$$\operatorname{supp}({\mu})=\{x\in X : \mu(U)>0\text{ for all $U$ open neighborhood of }x\}$$
But, in our case, for all $x \in X$, $\{x\}$ is an open neighborhood of $x$ and $\mu(\{x\})=0$. So, for all $x \in X$, $x \notin \operatorname{supp}({\mu})$. So, $\operatorname{supp}({\mu}) = \emptyset$.
Why is it odd to have $\operatorname{supp}({\mu}) = \emptyset$? Because, then we have  some $E\subseteq \operatorname{supp}({\mu})^c$, such that $\mu(E) >0$ (just take any uncountable set $E \subseteq X$).
Note that, for this example to work, it was key that any $E$ uncountable could not be written as countable union of the open sets $\{x\}$.
Remark:  The notion of support of measure is a minimal set outside of which the measure is null (similar to the notion of support of a function). It means that,
for any measurable set $E$, if $E \subseteq \operatorname{supp}({\mu})^c$ then $\mu(E)=0$.  However, using the definition of support of a measure as you posted, or as we can find here, this does not need to be the case for all  topological space, as we have seen in the example above,
Let us now prove that, if the topological space is second countable, then for any measurable set $E$, if $E \subseteq \operatorname{supp}({\mu})^c$ then $\mu(E)=0$.
Proof: Note that
$$\operatorname{supp}({\mu})^c=\{x\in X : \mu(U)=0\text{ for some $U$ open neighborhood of }x\}$$
A base of the topological space is a collection $B$ of open sets such that any open set $U \subseteq X$ is the union of elements (open sets) in $B$. The open sets in $B$ are called basic open sets.
So, if there is an open set $U$ such that $x\in U$ and $ \mu(U)=0$ then there is a basic open set $U'$ such that    $x\in U'\subseteq U$ and $\mu(U') =0$.
On the other, if there is a basic open set $U'$ such that    $x\in U'$ and $\mu(U') =0$, then we can take $U'$ as $U$, and we have that there is an open set $U$ such that $x\in U$ and $ \mu(U)=0$ . So, we have that: there is an open set $U$ such that $x\in U$ and $ \mu(U)=0$ if and only if  there is a basic open set $U'$ such that    $x\in U'$ and $\mu(U') =0$.
So, we have that
$$\operatorname{supp}({\mu})^c=\{x\in X : \mu(U')=0\text{ for some $U'$ basic  open set such that }x \in U'\}$$
So, for every $x \in \operatorname{supp}({\mu})^c$, there is a basic open set $U'$ such that $x\in U'$ and  $\mu(U')=0$. So
$$ \operatorname{supp}({\mu})^c \subseteq \bigcup_{U' \in B \textrm{ and }\mu(U')=0} U'\tag{1}$$
On the other hand, if $U'$ is a basic open set and $\mu(U')=0$, then for all $x\in U'$, $x \in \operatorname{supp}({\mu})^c$. So $U' \subseteq \operatorname{supp}({\mu})^c$. So,
$$ \bigcup_{U' \in B  \textrm{ and }\mu(U')=0} U' \subseteq \operatorname{supp}({\mu})^c \tag{2}$$
From $(1)$ and $(2)$,
$$ \operatorname{supp}({\mu})^c = \bigcup_{U' \in B \textrm{ and }\mu(U')=0} U'$$
So, it is clear that $\operatorname{supp}({\mu})^c$ is an open set.
Now, suppose that the topological space is second countable, that means, it has a countable base $B$. If $B$ is countable, we have
$$ \mu(\operatorname{supp}({\mu})^c) = \mu \left (\bigcup_{U' \in B \textrm{ and }\mu(U')=0} U' \right ) \leqslant \sum_{U' \in B \textrm{ and }\mu(U')=0}\mu(U') = 0$$
So, for any measurable set $E$, if $E \subseteq \operatorname{supp}({\mu})^c$ then $\mu(E)=0$.
