= Solution
To prove unboundedness, begin above any prescribed <ordinal> with $\gamma_0<\lambda$ and choose a strictly increasing sequence so that
$$
\gamma_{m+1}>\sup\bigcup_{\alpha<\gamma_m}B(\alpha).
$$
This upper bound is below $\lambda$: there are fewer than $\lambda$ countable sets in the union, and $\lambda$ is regular and uncountable. Put $\delta=\sup_m\gamma_m<\lambda$. For any $\alpha<\delta$, choose $m$ with $\alpha<\gamma_m$; then $B(\alpha)\subseteq\gamma_{m+1}\subseteq\delta$. Thus $\delta\in E$.
For closedness, suppose $\delta<\lambda$ is a limit point of $E$. Given $\alpha<\delta$, choose $\eta\in E$ with $\alpha<\eta<\delta$. Then $B(\alpha)\subseteq\eta\subseteq\delta$. Therefore $\delta\in E$ as well. Hence \b[$E$ is a <club set> in $\lambda$]. This is the <club of closure points for countable set-valued functions>.
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