Let $T$ be a normal bounded operator on a complex Hilbert space $H$. Suppose $\lambda$ is an isolated point in the spectrum of $T$. Then we can use functional calculus applied to the characteristic function $\mathbb{1}_{\{\lambda\}}$ to form the spectral projection $\mathbb{1}_{\{\lambda\}}(T)$. Since
$$(z-\lambda)\cdot\mathbb{1}_{\{\lambda\}}=0$$ as functions on the spectrum of $T$, it follows that for any $v\in H$,
$$(T-\lambda)\circ\mathbb{1}_{\{\lambda\}}(T)(v)=0,$$
so that the range of $\mathbb{1}_{\{\lambda\}}(T)$ lies within the $\lambda$-eigenspace of $T$.
Question: Is there a similarly simple argument that shows that $\mathbb{1}_{\{\lambda\}}(T)$ surjects onto the $\lambda$-eigenspace of $T$?
Remark: I am aware of this answer and this answer, but am looking for an alternative approach that does not involve spectral measures or Borel functional calculus.
Yes. If $Tx=\lambda x$, then $p(T)x=p(\lambda)x$ for any polynomial $p$. Taking norm limits, $f(T)x=f(\lambda)x$ for all continuous $f\in C(\sigma(T))$. As $\lambda$ is isolated, the function $1_{\{\lambda\}}$ is continuous on $\sigma(T)$. Thus $1_{\{\lambda\}}(T)x=x$.