As pointed out in my comment, an example would be given by the counting measures on $\mathbb Q \cap [0,1]$ and $(\sqrt{2} + \mathbb Q)\cap [0,1]$, respectively, on the compact metric space $X = [0,1]$.
Note that the same idea actually works for any compact metric space $X$ which has no isolated points.
Since you also asked about an example where the measure spaces are finite:
You can simply take "weighted measures", i.e. if $\{q_n\}_{n \in \mathbb N}$ is a enumeration of $\mathbb Q$, define a function
$$f(x) = \begin{cases} 2^{-n} & \text{if } x = q_n \\ 0 & \text{otherwise} \end{cases} $$
Now the weighted measure is given by $d\tilde \mu = f \, d\mu$, where $\mu$ is the counting measure on the rationals. This will then be finite
$$\int_\mathbb{R} \; d\tilde\mu = \int_\mathbb{R} f \; d\mu = \sum_{n = 1}^\infty 2^{-n} = 1$$
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).
Best Answer
The minimal (or smallest) closed set of full measure need not always be well-defined. The "example" (which we cannot show to exist in ZFC) is a discrete space $X$ with a probability measure $\mu$ on $\mathcal{P}(X)$ (look up real-valued measurable cardinals) that is $0$ on all singletons: for every closed set $C$ with $\mu(C)=1$, then if $x \in C$, $\mu(C\setminus\{x\}) =1$ too and this set is closed and strictly smaller. So only the empty set is minimal.
There are no points all of whose neighbourhoods are of positive measure. If course this measure is not tight, as the only compact sets are finite and thus measure $0$. So the intersection of all closed sets of full measure is $\emptyset$ which does not have full measure, and this is indeed the complement of the union all open null sets. So in that sense it's defined, but it's "useless".
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Added: Another classical, better, somewhat related example (in ZFC !) is the $0$-$1$-probability measure on $\omega_1$, which is explained here, the so-called Dieudonné measure (as Halmos calls it in in Measure Theory):
The set $X$ is just the ordinal number $\omega_1$ in its order topology (Munkres calls this space $W$, others call it $\Omega$) and a measure $\mu$ is defined on its Borel sets by $\mu(A) = 1$ iff there is a closed and unbounded set (a "club") that is a subset of $A$; otherwise $\mu(A) = 0$. As clubs are closed under countable intersections, this does define a somewhat unintuitive Borel measure on $X$, with the properties that $\mu(X) =1$, $\mu(\{x\}) = 0$ for all $x \in X$ (a singleton is not unbounded) and even $\mu(K) = 0$ for all compact subsets of $X$ (as these are bounded above and so don't contain a club). If $C$ is a closed set of measure $1$ (so it's actually a club) it will contain many isolated points $p$ (all successor ordinals are isolated in $X$) and then $C\setminus \{p\}$ is a strictly smaller closed set of full measure. So there cannot be a minimal closed set of full measure. Of course the measure is not inner regular for compact sets, but it is inner regular for closed sets, almost by definition. Ulam already showed that we cannot extend this measure to all subsets of $\omega_1$, but on the Borel sets it's well-defined.
If a probability Borel measure is tight (inner regular for compact sets) then defining $$N = \bigcup \{U| U \text{ open }, \mu(U)=0\}$$ the set $N$ has measure $0$: any compact $K \subseteq N$ is covered by finitely many open null sets, so has measure $0$ too and so $\mu(N) = \sup \{\mu(K): K \subseteq N, K \text{ compact }\} = 0$ and in that case the support
$$X\setminus \bigcup\{U: U \text{ open }, \mu(U)=0\}= \bigcap \{ C: C \text{ closed }, \mu(C)=1\}$$ has full measure and is indeed minimal with that property in that case.
So the “problem” is solved by considering suitably regular Borel measures. That’s why in analysis applications only Radon or tight measures are considered. See also the Handbook of Set-theoretic Topology (1984), one of my "bibles", in the chapter on Borel measures, for more discussion of such issues.