Internal energy for Phase Transitions
Can we justify that &quot;For sublimation of a solid at 1 atm  $\Delta U&gt;0$ at low temperature and $\Delta U&lt;0$ at high temperature?&quot;

No.
$\Delta U&gt;0$, always, for sublimation, because of the energy needed to separate the atoms or molecules in changing from the solid to the gas phase.
As for the enthalpy,
$$H = U +PV \Rightarrow \Delta H = \Delta U+\Delta (PV)=\Delta U+ P\Delta V+ V \Delta P$$
Thus at constant pressure:
$$\Delta H = \Delta U+ P\Delta V \approx  \Delta U+ nRT$$
[Here I've used the ideal gas law to approximate the volume of the gas, and ignored the volume of the solid, which (at 1 atm, room temperature) is about 3 orders of magnitude smaller.]
I.e., $\Delta H$ will be even more positive than $\Delta U$, because of the pV-work required to make space for the gas.
The reason why sublimation becomes more favorable as the temperature increases is because sublimation has a positive $\Delta S$, and $\Delta G = \Delta H - T \Delta S$.  But, irrespective of temperature, at any temperature and pressure at which the solid can exist,  $\Delta U_{solid-&gt;gas} &gt;0 $.
$\Delta U &gt;0$ for liquid-&gt;gas and solid-&gt;liquid phase transitions as well (with the possible exception of the solid-&gt;liquid phase transitions for helium-3 and helium-4 at extremely low temperatures which, at least according to https://en.wikipedia.org/wiki/Enthalpy_of_fusion, have $\Delta H &lt; 0$; but solid helium doesn't exist at the 1 atm pressure specified by the OP).
*Yes, you might be able to come up some extreme hypothetical mechanical system where the solid is under such great pressure that the intermolecular forces are so far into the repulsive part of their potentials that $\Delta U_{solid-&gt;gas} &lt;0 $.  But that's clearly not what the OP had in mind, since s/he specified 1 atm.  And, in addition, such an arrangement would require that only the solid, but not the gas, be at that extreme pressure.  So, strictly speaking, that $\Delta U$ would not be for the sublimation alone, it would be for the sublimation plus the pressure change.