In reading through some notes on symmetric spaces, I came across the statement that the group of isometries of a complete Riamannian maifold forms a lie group. The notes referenced a book by Kobayashi, although the proof there used principal bundles, which I am not all too familiar with.
From the definition given in Spivak, a function $f:M\to N$ is an isometry if if is smooth, injective and satisfies $f^*(\langle\cdot,\cdot\rangle_N) = \langle\cdot,\cdot\rangle_M$.
Defining an isometry from a smooth manifold $M$ to itself as a bijective isometry, it follows directly that the group of isometries is indeed a group (as $f^*(\langle\cdot,\cdot\rangle) = \langle\cdot,\cdot\rangle \implies \langle\cdot,\cdot\rangle = (f^{-1})^*(\langle\cdot,\cdot\rangle)$ ).
Thus, it seems that the only thing left to do is to show that is itself smooth. Is there a way to do this without having to refer to advanced constructions such as principle bundles?