Is the composition of a strictly concave function with a concave function strictly concave?
Solution 1:
A simpler counterexample, which helps shed some light: $g$ can be a constant function.
The issue in your argument is that you implicitly assume $g(x)\neq g(y)$ to obtain the strict inequality.
In you take $g(x) = x_1+x_2$, then you can have $g(x)=g(y)$ even if $x\neq y$: take for instance $x=(x_1,x_2)$ and $y=(x_2,x_1)$. You need injectivity somewhere for your argument to go through.