I will concentrate on comparing (3) and (4). The definition (1) is meant for finite signed measures, whereas all the other definitions are meant for arbitrary positive measures; (1) is equivalent to (4) in the case of finite positive measures. (2) appears to be equivalent to (4) ["locally finite" can mean "finite on compact sets", although it is sometimes taken to mean "finite on the elements of some topological basis"; these are equivalent in the LCH (locally compact Hausdorff) case]. Finally, (5) does not appear to be a definition at all, but rather a description of a definition.
Now then,
i) In the case of a second countable LCH space, every locally finite measure satisfies both (3) and (4) (Theorem 7.8 of [1]). This is the most commonly considered scenario in applications, which is why almost no one bothers to carefully sort out the differences between the different definitions.
ii) In the case of a sigma-compact LCH space, (3) and (4) are equivalent. The forward direction is Corollary 7.6 of [1]; the backwards direction follows from the forward direction together with (iv) below (but I'm sure there is an easier proof).
iii) (3) and (4) are not equivalent in general, even for LCH metrizable spaces (Exercise 7.12 of [1]).
iv) In an LCH space, there is a bijection between
A) measures satisfying (3),
B) measures satisfying (4), and
C) positive linear functionals on the space of continuous functions with compact support.
(The Riesz representation theorem gives either (A)<->(C) or (B)<->(C), depending on where you look; (A)<->(B) is in the Schwarz book mentioned by Joe Lucke; see also Exercise 7.14 of [1])
[1] G. B. Folland, Real Analysis: Modern Techniques and Their Applications
Note: In [1], "Radon" refers to measures satisfying (3).
For an example with a probability measure, consider the following standard counterexample: let $X = [0, \omega_1]$ be the uncountable ordinal space (with endpoint), with its order topology. This is a compact Hausdorff space which is not metrizable. Define a probability measure on the Borel sets of $X$ by taking $\mu(B) = 1$ if $B$ contains a closed uncountable set, $\mu(B)=0$ otherwise. It is known that this defines a countably additive probability measure; see Example 7.1.3 of Bogachev's Measure Theory for the details.
If $x \in X$ and $x < \omega_1$, then $[0, x+1)$ is an open neighborhood of $x$ which is countable, hence has measure zero. So $x$ is not in the support of $\mu$. In fact the support of $\mu$ is simply $\{\omega_1\}$. But $\mu(\{\omega_1\}) = 0$.
Best Answer
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 $\mu(E)=0$.