 Solution

For regular $\kappa$, the cobounded filter on a regular cardinal is $\kappa$-complete. Part (c) extends it to a $\kappa$-complete ultrafilter $U$. Since $\kappa \setminus \{\alpha \}$ is cobounded for every $\alpha <\kappa$, no singleton belongs to $U$; hence $U$ is nonprincipal. Thus every strongly compact cardinal is measurable.