Which of the following compounds has high lattice enthalpy?

Lattice enthalpy depends on ionic charge magnitude and the sum of ionic radii, quantified by the relation $\text{Lattice Enthalpy} \propto \frac{|q_1 q_2|}{r_+ + r_-}$.

Among LiCl, NaCl, BeF$_2$, and CaCl$_2$, compounds with doubly charged cations (Be$^{2+}$ or Ca$^{2+}$) exhibit higher lattice enthalpy due to increased charge product. BeF$_2$ and CaCl$_2$ both have charge product $|(+2) \times (-1)| = 2$, whereas LiCl and NaCl have charge product $1$.

Be$^{2+}$ ($\approx 27$ pm) is smaller than Ca$^{2+}$ ($\approx 100$ pm), and F$^-$ ($\approx 133$ pm) is smaller than Cl$^-$ ($\approx 181$ pm), resulting in a notably smaller sum of radii for BeF$_2$ ($\approx 160$ pm) compared to CaCl$_2$ ($\approx 281$ pm).

Approximating the proportionality: for BeF$_2$, $\frac{2}{160} = \frac{1}{80}$; for LiCl, $\frac{1}{257}$. The value for BeF$_2$ is significantly larger, confirming it has the highest lattice enthalpy.

The answer is $\boxed{\text{C}}$.