Yes. It's more generally true for every faithful $C^0$ unitary representation $\pi$ of $G$.
Indeed if this is not a homeomorphism onto its image, there exists an ultrafilter $\eta$ on $G$, not converging to $1$, such that $\lim_{g\to\eta}\lambda_g=\mathrm{id}$ (for the strong topology, i.e., $\lim_{g\to\eta}\pi(g)v=v$ for every $v$ in the Hilbert space.
If $\eta$ is unbounded (i.e. no compact subset of $G$ is in $\eta$), then we get a contradiction, since the $C^0$ property implies $\lim_{g\to\eta}\langle\pi(g)v,v\rangle=0$ for every $v$ in the Hilbert space.
If $\eta$ is bounded, then $\eta$ has a limit $g_0$ (which by assumption is not $1_G$), and we deduce $\lim_{g\to\eta}\pi(g)=\pi(g_0)$. So $\pi(g_0)=\mathrm{id}$, contradicting the faithfulness assumption.