Vitali-type set with given outer measure
One can proceed directly with the Vitali construction, without need for any scaling.
Namely, just carry out the Vitali construction, but ensure that the resulting Vitali set is contained in the interval $[0,a]$. That is, declare that two reals are equivalent if their difference is rational, and observe that every real is equivalent to a real in the interval $[0,a]$. Let $V\subset [0,a]$ select exactly one element from each equivalence class. Observe that the rational translations $V+q$, working modulo $1$ so as to regard $V+q\subset [0,1]$, are disjoint and union up to the whole interval $[0,1]$. It follows easily that $V$ is not measurable and has inner measure $0$, since otherwise the translates modulo $1$ would have infinite measure inside $[0,1]$, which is impossible. Thus, the complement $[0,a]-V$ is non-measurable and has outer measure $a$, as desired.
I would like to note that this argument shows that the outer measure of the classical Vitali set is not determined by the usual features of that set. For example, if all you know about a Vitali set $V$ is that it is contained in $[0,1]$ and contains exactly one element of each equivalence class, then it follows that $V$ is non-measurable and has inner measure $0$, but for all you know, $V$ is actually contained in a very tiny interval $[0,\epsilon]$, and could have tiny outer measure.
This was my answer to the question as originally phrased: Once you have one nonmeasurable set with finite outer measure, you can get all positive values by rescaling.
With the additional requirement that the set is contained in $[0,1]$, this argument will work as long as you have a nonmeasurable subset of outer measure $1$. Sierpiński and Lusin showed that $[0,1]$ can be decomposed into uncountably many nonmeasurable sets of outer measure $1$ in an article published in 1917, but I do not know the details. Robert Israel sketches a construction of a Vitali subset of [0,1] with outer measure 1 in a 1997 sci.math post.
(Oh, and strictly speaking the answer is no, you can't have $x=0$; I suppose you meant $x\in(0,1]$.)
As for the axiom of choice, I'm no expert but have been told that it is consistent with ZF that all subsets of the real line are Lebesgue measurable.
Peter LeFanu Lumsdaine points out in a comment that the result is due to Robert Solovay. The 1970 article is on JSTOR, and begins as follows:
