Martin Greinecker's answer (in a comment) is correct in the context of the usual axioms of set theory, including the axiom of choice. If one works in set theory without the axiom of choice, then one cannot prove that $\mathcal P(\mathbb R)$ admits any total order at all (even without the requirement that it extend $\subseteq$). On the other hand, the $\subseteq$ ordering of $\mathcal P(\mathbb N)$ can be explicitly extended to a total ordering, given by lexicogaphically ordering the characteristic functions of subsets of $\mathbb N$.

Michael Greinecker's answer (in a comment) is correct in the context of the usual axioms of set theory, including the axiom of choice. If one works in set theory without the axiom of choice, then one cannot prove that $\mathcal P(\mathbb R)$ admits any total order at all (even without the requirement that it extend $\subseteq$). On the other hand, the $\subseteq$ ordering of $\mathcal P(\mathbb N)$ can be explicitly extended to a total ordering, given by lexicogaphically ordering the characteristic functions of subsets of $\mathbb N$.