Brezis' exercise 5.20.3: an approach based on weak topology

Question

I'm trying to fill in the details of the proof of part (3.) of of exercise 5.20 in Brezis' Functional Analysis.

Let $(H, \langle \cdot, \cdot \rangle)$ be a real Hilbert space and $|\cdot|$ its induced norm. Let $S :H \to H$ be a bounded linear operator such that $\langle Su, u \rangle \ge 0$ for all $u \in H$. Let $N(S)$ be the kernel of $S$ and $R(S)$ the range of $S$. Then

1. $N(S) = R(S)^\perp$.
2. The map $I + t S$ is bijective for all $t>0$. Here $I:H \to H$ is the identity map.
3. $\lim_{n \to \infty} (I +n S)^{-1} f = \pi_{N(S)} f$ for all $f\in H$. Here $\pi_{N(S)} f$ is the orthogonal projection of $f$ onto $N(S)$.

Could you verify my below attempt? Thank you so much for your help!

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Let $x_n := (I +n S)^{-1} f$. Then $x_n + n S x_n =f$. Then $$ |x_n|^2 + 2n \langle x_n, Sx_n \rangle +n^2|Sx_n|^2 = |f|^2. $$

Because $\langle x_n, Sx_n \rangle \ge 0$, we get $|x_n| \le |f|$ and $n|Sx_n| \le |f|$ for all $n$. It suffices to prove that each subsequence of $(x_n)$ has a further subsequence that converges to $\pi_{N(S)} f$. We take a subsequence of $(x_n)$ and for ease of notation we still denote this subsequence by $(x_n)$.

Because $(x_n)$ is bounded, there is a subsequence $\varphi$ of $\mathbb N$ and $x \in H$ such that $x_{\varphi (n)} \to x$ in the weak topology $\sigma(H, H^*)$. It remains to prove that $x = \pi_{N(S)} f$ and that $x_{\varphi (n)} \to x$ in norm topology. To simplify notations again, we assume WLOG $\varphi(n)=n$ for all $n$. We have $n|Sx_n| \le |f|$ for all $n$, so $Sx_n \to 0$. On the other hand, $Sx_n \to Sx$. Then $Sx=0$ and thus $x \in N(S)$.

Let's prove that $x = \pi_{N(S)} f$. Let $v \in N(S)$. It suffices to prove $\langle f-x, v \rangle=0$. By (1.), we have $\langle f-x_n, v \rangle = n\langle Sx_n, v \rangle = 0$. Then $\langle f-x, v \rangle = \lim_n \langle f-x_n, v \rangle = 0$.

To prove that $x_n \to x$ in norm topology, it suffices to prove that $\limsup_n |x_n|^2 \le |x|^2$. We have $$ \langle n Sx_n, x_n \rangle = \langle f-x_n, x_n \rangle \ge 0. $$

So $|x_n|^2 \le \langle f, x_n \rangle$. Then $\limsup_n |x_n|^2 \le \langle f, x \rangle = |x|^2$. This completes the proof.