Please note: This question has been edited after it became clear from Christian Remling's answer that the original formulation was far from what I really meant to ask.

Let $T\ne 0$ be a self-adjoint operator on a Hilbert space $H$, with spectrum $\sigma(T)$. For any $x∈H$, denote by $μ_x$ the spectral measure of $T$ at $x$, that is the unique Borel measure on $\sigma(T)$ such that

$$ ⟨x,f(T)x⟩ = \int_{\sigma(T)} f(λ) d \mu_{x}(λ) \quad \forall f \in \mathcal{C}(\sigma(T),\mathbb{C}). $$

Then, one can prove that $$ \overline{\bigcup_{x\in X} Supp(\mu_x)}=\sigma(T)$$ for any orthonormal basis $X$ of $H$.

Suppose that there exists a closed subspace $H_0$ of $H$ and a positive integer $k$ such that $T|_{T^i H_0}$ is injective for all $i=0,\dots,k-1$, and $$ H=H_0 \overset{\perp}{\oplus} TH_0 \overset{\perp}{\oplus} \dots \overset{\perp}{\oplus} T^{k-1} H_0,\qquad T^{k}H_0=H_0.$$

If $k=1$, then $T$ is invertible, hence $0\in \sigma(T)$ if and only if the inverse $T^{-1}$ is unbounded. Therefore, either $0\not\in \sigma(T)$ or $\lambda=0$ is an accumulation point of $\sigma(T)$.

Suppose now that $k\ge 2$, and suppose that there is an orthonormal basis $X_0$ of $H_0$ such that $\mu_x(\{0\})>0$ for all $x\in X_0$. Then, one has $0\in \sigma(T)$. Under some additional assumptions I can actually show that $\lambda=0$ must be an isolated eigenvalue. However, I was actually wondering whether that might always be the case.

Addendum: This is how I would like reason. Let $$ X:=X_0\cup TX_0 \cup \dots \cup T^{k-1}X_0. $$

Let $\mathfrak{M}$ be the set of discrete positive measures $\nu$ on $X$ with the following property: $$\forall A\subseteq X,\; \nu(A)=0\implies \nu(TA)=0.$$

Then, for all non-zero $\nu\in\mathfrak{M}$, one has $\nu(X_0)>0$. Hence $$ \int_{X} \mu_{x}(\{0\})d\nu(x) \ge \int_{X_0} \mu_{x}(\{0\})d\nu(x)>0.$$

It seems to me that this could be enough to conclude that $\lambda=0$ must be an isolated eigenvalue. However, I can only show this with extra assumptions.