The book Probability measures on metric spaces by K. R. Parthasarathy is my standard reference; it contains a large subset of the material in Convergence of probability measures by Billingsley, but is much cheaper! Parthasarathy shows that every finite Borel measure on a metric space is regular (p.27), and every finite Borel measure on a complete separable metric space, or on any Borel subset thereof, is tight (p.29). Tightness tends to fail when separability is removed, although I don't know any examples offhand.
(Definitions used in Parthasarathy's book: $\mu$ is regular if for every measurable set $A$, $\mu(A)$ equals the supremum of the measures of closed subsets of $A$ and the infimum of open supersets of $A$. We call $\mu$ tight if $\mu(A)$ is always equal to the supremum of the measures of compact subsets of $A$. Some other texts use "regular" to mean "regular and tight", so there is some room for confusion here.)
I think I have been able to reproduce the "argument by Wagoner" (perhaps it was removed from the published version?). It certainly holds in more generality that what I have written below, using the notion of "direct sum group" in Wagoner's paper (which unfortunately seems to be a little mangled).
Let $M$ be a homotopy commutative topological monoid with $\pi_0(M)=\mathbb{N}$. Choose a point $1 \in M$ in the correct component and let $n \in M$ be the $n$-fold product of 1 with itself, and define $G_n = \pi_1(M,n)$. The monoid structure defines homomorphisms
$$\mu_{n,m} : G_n \times G_m \longrightarrow G_{n+m}$$
which satisfy the obvious associativity condition. Let $\tau : G_n \times G_m \to G_m \times G_n$ be the flip, and
$$\mu_{m,n} \circ \tau : G_n \times G_m \longrightarrow G_{n+m}$$
be the opposite multiplication. Homotopy commutativity of the monoid $M$ not not ensure that these two multiplications are equal, but it ensures that there exists an element $c_{n,m} \in G_{n+m}$ such that
$$c_{n,m}^{-1} \cdot \mu_{n,m}(-) \cdot c_{n,m} = \mu_{m,n} \circ \tau(-).$$
Let $G_\infty$ be the direct limit of the system $\cdots \to G_n \overset{\mu_{n,1}(-,e)}\to G_{n+1} \overset{\mu_{n+1,1}(-,e)}\to G_{n+2} \to \cdots$.
Theorem: the derived subgroup of $G_\infty$ is perfect.
Proof: Let $a, b \in G_n$ and consider $[a,b] \in G'_\infty$. Let me write $a \otimes b$ for $\mu_{n,m}(a, b)$ when $a \in G_n$ and $b \in G_m$, for ease of notation, and $e_n$ for the unit of $G_n$.
In the direct limit we identify $a$ with $a \otimes e_n$ and $b$ with $b \otimes e_n$, and we have
$$b \otimes e_n = c_{n,n}^{-1} (e_n \otimes b) c_{n,n}$$
so $b \otimes e_n = [c_{n,n}^{-1}, (e_n \otimes b)] (e_n \otimes b)$. Thus
$$[a \otimes e_n, b \otimes e_n] = [a \otimes e_n, [c_{n,n}^{-1}, (e_n \otimes b)] (e_n \otimes b)]$$
and because $e_n \otimes b$ commutes with $a \otimes e_n$ this simplifies to
$$[a \otimes e_n, [c_{n,n}^{-1}, (e_n \otimes b)]].$$
We now identify this with
$$[a \otimes e_{3n}, [c_{n,n}^{-1}, (e_n \otimes b)] \otimes e_{2n}]$$
and note that $a \otimes e_{3n} = c_{2n,2n}^{-1}(e_{2n} \otimes a \otimes e_{n})c_{2n,2n} = [c_{2n,2n}^{-1}, (e_{2n} \otimes a \otimes e_{n})]\cdot (e_{2n} \otimes a \otimes e_{n})$.
Again, as $(e_{2n} \otimes a \otimes e_{n})$ commutes with $[c_{n,n}^{-1}, (e_n \otimes b)] \otimes e_{2n}$ the whole thing becomes
$$[a,b]=[[c_{2n,2n}^{-1}, (e_{2n} \otimes a \otimes e_{n})], [c_{n,n}^{-1}, (e_n \otimes b)] \otimes e_{2n}],$$
a commutator of commutators.
Best Answer
A counterexample is shown on the cover of the paperback edition of the classic textbook Homology Theory by Hilton and Wylie. This can be viewed on the amazon webpage for the book. The example consists of the wedge of two copies of a cone, the cone on the sequence 1/2, 1/3, 1/4, ... together with its limit point 0. With one choice of basepoints the wedge is not contractible, but with other choices it is.