I'm asked to determine \$ V_x, I_x \$ in this circuit: 
First, from first look the middle "mesh" is incomplete so there is no complete path for I_x to return thus KCL requirements aren't fulfilled, so
\$ I_x = 0 A \$
Second, we can work on the two other meshes independently, \$ 8 = (4+2)i_1 \iff i_1 = 1.333 A \$ and \$ 4 = (6+3)i_2 \iff i_2 = \frac{4}{9} A \$ thus \$ V_x = - 2.667 V \$
Is this logic correct or is there a flaw?
Indeed, there is no current I_x because there is no return path. So you can solve each rectangle (ie each group of a voltage source and 2 resistors) separately.
The only impact of having this resistor (or a wire, it would be the same) connecting the 2 rectangles is that it enables to define the voltage between 2 nodes, one in each rectangle, if you wish (but you are not asked to in this exercise).
So either this exercise is a "trap" (which you successfully detected) to test if you understand that there is no current I_x.
Or It could be an error in the exercise (for example a missing wire at the bottom)
Or it could be the beginning of a bigger exercise, with some return path added later on.
For the computations of Vx, it seems correct to me (but I don't see why you need to compute i2)
