The proof that $C$ is closed is straightforward. Suppose that $\langle\alpha_n:n\in\omega\rangle$ is a strictly increasing sequence in $C$, and let $\alpha=\sup\{\alpha_n:n\in\omega\}$ Then $j[\alpha\times\alpha]=j\left[\bigcup_{n\in\omega}\alpha_n\times\alpha_n\right]=\bigcup_{n\in\omega}j[\alpha_n\times\alpha_n]=\bigcup_{n\in\omega}\alpha_n=\alpha\;,$ so $\alpha\in C$, and $C$ is closed.
One of the standard approaches to showing that a set is unbounded works here. Start with any $\alpha_0\in\omega_1$; you want to find an $\alpha\ge\alpha_0$ such that $j[\alpha\times\alpha]=\alpha$. If $j[\alpha_0\times\alpha_0]=\alpha_0$, you’re done, of course, but you can’t count on that If $j[\alpha_0\times\alpha_0]\nsubseteq\alpha_0$, let $\alpha_1=\min\{\alpha\in\omega_1:\alpha_1>\alpha_0\land\alpha_0\subseteq j[\alpha_0\times\alpha_0]\subseteq\alpha\}\;.$ Again, if $j[\alpha_1\times\alpha_1]=\alpha_1$, you’re done, but that’s too much to hope for. In fact, we’ll just ignore the possibility and pretend that it doesn’t happen; as you’ll see, there’s no harm in that. For each $n\in\omega$ simply let $\alpha_{n+1}=\min\{\alpha\in\omega_1:\alpha_{n+1}>\alpha_n\land\alpha_n\subseteq j[\alpha_n\times\alpha_n]\subseteq\alpha\}\;.$ The result is a strictly increasing sequence $\langle\alpha_n:n\in\omega\rangle$ in $\omega_1$. Let $\alpha$ be the supremum of this sequence. Then $j[\alpha\times\alpha]=\bigcup_{n\in\omega}j[\alpha_n\times\alpha_n]\subseteq\bigcup_{n\in\omega}\alpha_{n+1}=\alpha=\bigcup_{n\in\omega}\alpha_n\subseteq\bigcup_{n\in\omega}j[\alpha_{n+1}\times\alpha_{n+1}]=j[\alpha\times\alpha]\;,$ so $\alpha\in C$.