Suppose m x </msub> , M x </msub> , m y </

No, the solution is not unique, unless you have special configurations of the given bounds. Also, this is more of an exercise in basic real analysis than linaer algebra.
If a solution exists at all, we have
$2m_x\le <!--\; \le \; -->x_l+x_h\le <!--\; \le \; -->2M_x$
and
$2m_y\le <!--\; \le \; -->y_l+y_h\le <!--\; \le \; -->2M_y$
and hence it must be true that
$S=[2m_x,2M_x]\cap <!--\; \cap \; -->[2m_y,2M_y]\ne <!--\; \ne \; -->\mathrm{\varnothing <!--\; \varnothing \; -->}.$
I'll ignore the degenerate cases where $M_x\le <!--\; \le \; -->m_x$ and/or $M_y\le <!--\; \le \; -->m_y$, they don't bring anything new to the table (and fall under the special configurations I mentioned above), so I assume that the intervals whose intersection S is are proper intervals with positive length.
$S$ is an interval, but it might be degenerate and just a point. Those are the 2 cases $2m_x<2M_x=2m_y<2My$ and $2m_y<2M_y=2m_x<2Mx$. In the first case, it must be that $x_l+x_h=2Mx=2m_y=y_l+y_h$ and thus $x_l=x_h=M_x=m_y=y_l=y_h$. The second case has similiar consequences, again we get $x_l=x_h=y_l=y_h$.
But again, this is a special configuration of the given bounds.
If $|S|>1$, then $S$ is a proper interval $S=[a,b]$ with $a<b$. In this case, there are infinitely many $s\in <!--\; \in \; -->(a,b)$ and for each of them $x_l=x_h=y_l=y_h=\frac{s}{2}$ is a solution. That's because
$s\in <!--\; \in \; -->(a,b)\subset <!--\; \subset \; -->[a,b]=S\subseteq <!--\; \subseteq \; -->[2m_x,2M_x]$
and hence
$m_x\le <!--\; \le \; -->\frac{s}{2}\le <!--\; \le \; -->M_x,$
and the same is true for $m_y/M_y$.
Moreover, because we chose $s$ from the open interval $(a,b)$, we do not have equality in either side of the inequality above, so we get
$m_x<\frac{s}{2}<M_x$
That means $\frac{s}{2}$ is an inner point of the interval $[m_x,M_y]$ and so ther exists a ${\mathrm{\delta <!--\; \delta \; -->}}_s>0$ such that
$m_x\le <!--\; \le \; -->\frac{s}{2}-<!--\; -\; -->{\mathrm{\delta <!--\; \delta \; -->}}_s<\frac{s}{2}+{\mathrm{\delta <!--\; \delta \; -->}}_s\le <!--\; \le \; -->M_x.$
Now that means in the above solution $x_l=x_h=y_l=y_h=\frac{s}{2}$ we can replace $x_l,x_h$ with infintely many other pairs
$x_l=\frac{s}{2}-<!--\; -\; -->\mathrm{\delta <!--\; \delta \; -->},x_h=\frac{s}{2}+\mathrm{\delta <!--\; \delta \; -->}$
for any $0<\mathrm{\delta <!--\; \delta \; -->}\le <!--\; \le \; -->{\mathrm{\delta <!--\; \delta \; -->}}_s$
Absolutely the same argument can be made for $y_l,y_h$
So unless you have initial bounds $m_x,M_x,m_y,M_y$ that trivially force your unknowns to be either bound of the interval, you will have infinitely many solutions, also with infinitely many of them such that $x_l\ne <!--\; \ne \; -->x_h$ and $y_l\ne <!--\; \ne \; -->y_h$.
To me it looks like what happens often when dealing with inequalities. You come to a point where you see something that would have nice consequences if it could be proven. But unfortunately, your system does generally not have a unique solution and it is not even close.