Showing $\{h + Th: h \in H\}$ is closed under $T = T^* = T^{-1}$. This is a part of a question (2.21) from Elementary Functional Analysis by Barbara D.MacCluer (GTM 253).
Suppose that $T: H \rightarrow H$ is an operator on a Hilbert space $H$ such that $T = T^{-1}$ and $T = T^* $, where $T^*$ is the adjoint operator of $T$. Show that the sets $$H_1 = \{h + Th: h \in H\}$$ and $$H_2 = \{h - Th: h \in H\}$$ are closed subspaces of $H$ with $H = H_1 \oplus H_2$, and the restriction of $T$ to $H_1$ is the identity $I$ while the restriction of $T$ to $H_2$ is $-I$.
Conversely, show that if $H$ is the direct sum of two subspaces $H_1$ and $H_2$ with $T(h) = h$ for $h \in H_1$ and $T(h) = -h$ for $h \in H_2$, then $T = T^{-1}$ and $T^* = T$.

I would appreciate it if my reasoning for $H_1$ is closed can be verified. Suppose $h_n + Th_n \rightarrow h \in H$. Then, $$h + Th = \lim (h_n + Th_n) + T\lim (h_n + Th_n) = \lim(2h_n + 2Th_n) = 2h.$$ That means, $h_n + Th_n \rightarrow h = h/2 + T(h/2) \in H$.
If this seems to be correct, where did I use the condition that $T = T^*$? Did I miss something?

The question didn't explicitly specify that $T$ was continuous, so how to you know that $T \lim_n (h_n + T h_n) = \lim_n (h_n + T h_n)$?
In fact, continuity of $T$ follows from the fact that $T^* = T^{-1}$ so that $T$ is an isometry since $\langle T x, Ty\rangle = \langle T^*T x, y\rangle = \langle x, y\rangle$.
