Just use the definitions.

Consider first the case of an arbitrary filtration $\mathcal F_t$ (not necessarily right-continuous). For every $s$, the random variable $\lambda(s)$ is a Markov time with respect to $\mathcal F_{t+}$, in the sense that $$\{\lambda(s) < t\} \in \mathcal F_t \quad \text{for every $t$.}$$ By definition, $E \in \mathcal F_{\lambda(s)+}$ if and only if $$E \in \mathcal F_\infty \quad \text{and} \quad E \cap \{\lambda(s) < t\} \in \mathcal F_t \quad \text{for every $t$.}$$ Similarly, $E \in \mathcal F_{\lambda(s+\varepsilon)+}$ if and only if $$E \in \mathcal F_\infty \quad \text{and} \quad E \cap \{\lambda(s + \varepsilon) < t\} \in \mathcal F_t \quad \text{for every $t$.}$$ It follows that if $E \in \bigcap_{\varepsilon > 0} \mathcal F_{\lambda(s+\varepsilon)+}$, then $E \in \mathcal F_\infty$ and $$E \cap \{\lambda(s) < t\} = \bigcup_n E \cap \{\lambda(s + \tfrac{1}{n}) < t\} \in \mathcal F_t$$ for every $t$, and thus $E \in \mathcal F_{\lambda(s)+}$ (here indeed we use right-continuity of $\lambda$). Of course, $\bigcap_{\varepsilon > 0} \mathcal F_{\lambda(s+\varepsilon)+}$ contains $\mathcal F_{\lambda(s)+}$, and consequently the two $\sigma$-algebras are equal.

If $\mathcal F_t$ is right-continuous, then $\mathcal F_{t+} = \mathcal{F}_t$, and consequently $\lambda(s)$ is a Markov time with respect to $\mathcal F_t$ (in the sense that $\{\lambda(s) \leqslant t\} \in \mathcal F_t$), and $\mathcal F_{\lambda(s)+} = \mathcal F_{\lambda(s)}$. Here $E \in \mathcal F_{\lambda(s)}$ if and only if $$E \in \mathcal F_\infty \quad \text{and} \quad E \cap \{\lambda(s) \leqslant t\} \in \mathcal F_t \quad \text{for every $t$.}$$ It follows that indeed $\mathcal G_s = \mathcal F_{\lambda(s)}$ is right-continuous.