How can we prove the inequality TdS > dU + pdV?
For a closed system with a single phase of a pure substance, when only expansion work is possible
$$\mathrm dU= -\mathrm p\,\mathrm dV + T\,\mathrm dS, \label{eqn:1}\tag{1}$$
where $p$ (the state variable) is the internal pressure and $T$ (the state variable) is the internal temperature of the system.
The first law, however, says more generally that
$$\mathrm dU = \mathrm dw + \mathrm dq. \label{eqn:2}\tag{2}$$
Equation \eqref{eqn:2} was derived by applying equation \eqref{eqn:1} to a reversible process.
For the reversible process, thermal and mechanical equilibrium between internal and external pressure means that $p_\mathrm{ext} = p$ and $T_\mathrm{ext} = T,$ and, combined with the second law, that
$$\mathrm dw = -p_\mathrm{ext}\,\mathrm dV = -p\,\mathrm dV \tag{3}$$
and
$$\mathrm dq = -T_\mathrm{ext}\,\mathrm dS_\mathrm{ext} = T\,\mathrm dS. \tag{4}$$
More generally, the second law says that
$$\mathrm dS_\mathrm{tot} = \mathrm dS_\mathrm{sys} + \mathrm dS_{surr} \ge 0 \tag{5}$$
the equality holding for a reversible process. Dropping the label for the system
$$\mathrm dS \ge -\mathrm dS_\mathrm{surr} \quad\leftrightarrow\quad \mathrm dS \ge \frac{\mathrm dq}{T_\mathrm{surr}}. \tag{6}$$
Therefore for an isothermal process (thermal equilibrium condition applies)
$$\mathrm dS \ge \frac{\mathrm dq}{T}. \tag{7}$$
The inequality applies for instance to an isothermal free expansion.