How to prove that a left-invariant metric on a Lie group is smooth?

A natural approach is to show that the map $f(x,v)= (dL_{x^{-1}})_x(v)$ is smooth: The map $F:G\times G \to G$ given by $F(x,y)=xy$ satisfies $dF((x,u),(y,v))=(dL_{x})_y(v)+(dR_y)_x(u)$. Embed $TG$ in $T(G\times G)\cong TG \times TG$ via $(x,v)\mapsto ((x^{-1},0),(x,v))$. The composition $TG \hookrightarrow T(G \times G) \to TG$ is smooth and gives the desired map $$(x,v)\mapsto ((x^{-1},0),(x,v))\mapsto (dL_{x^{-1}})_x (v).$$

Now, given vector fields $X$ and $Y$, we have a smooth map $G \to \mathbb{R}$ given by $x\mapsto \langle X(x),Y(x)\rangle_x=\langle \phi\circ X(x), \phi \circ Y(x)\rangle_e$. (Note that restricting the codomain of the composition from $TG$ down to $T_e G$ results in a smooth map because $T_e G$ is an embedded submanifold of $TG$; see the comments below.)

I think it is just the composition of smooth maps. Let's try to state this formaly.

Consider the "double-tangent bundle" (this is not an official notation, I'm not sure it have a standard name) $T^2G=\bigsqcup_{x\in G}T_x G \times T_x G$ and $\pi : T^2G \rightarrow G$ the canonical projection.

Then, we can define on $T^2G$ the map (with $\mathfrak{g}=T_eG$ the Lie algebra) : $$ \begin{array}{rccc} L : & T^2G & \longrightarrow & \mathfrak{g} \times \mathfrak{g}\\ & (u,v) & \longmapsto & \left( (dL_{\pi(u,v)^{-1}})_{\pi(u,v)} u, (dL_{\pi(u,v)^{-1}})_{\pi(u,v)} v\right), \end{array}$$

Because $\pi$ and $(x,y) \mapsto L_x(y)$ are smooth, so is $L$, by composition of smooth functions. The left-invariant metric is then just the composition of $L$ with the metric on $\mathfrak{g}$ and thus is also smooth, seen as a map from $T^2G$ to $\mathbb{R}$.