Let $A\colon D(A) \to \mathcal H$ be a self-adjoint operator on a separable Hilbert space $\mathcal H$. Prove that $$\sigma_{\text{ess}}(A^2) = \left( \sigma_{\text{ess}}(A) \right)^2,$$where $\sigma_{\text{ess}}(A^2)$ denotes the essential spectrum, that is $$\sigma_{\text{ess}}(A) := \\ \sigma(A) \setminus \{\lambda \in \sigma(A): \lambda \text{ is an eigenvalue of } A \text{ isolated in } \sigma(A) \text{ and has finite mult.}\}$$
I attempted this using Weyl sequences: Let $\{x_n\}_{n\in \mathbb N}$ be a (singular) Weyl sequence, that is $\lVert x_n \rVert = 1, \lVert (A-\lambda) x_n \rVert \to 0$ strongly, $x_n \rightharpoonup 0$ weakly. I want to show that $-\sqrt{\lambda}$ or $\sqrt{\lambda}$ is in the essential spectrum of $A$. To that end, I think I can claim that $\{x_n\}_{n\in \mathbb N}$ is a singular Weyl sequence for $-\sqrt{\lambda}$ or, if not, $\left((A+\sqrt{\lambda})x_n\right)_{n\in \mathbb N}$ is a Weyl sequence for $\sqrt{\lambda}$. However, this ended in messy calculations and I couldn't conclude without knowing $A$ to be bounded. Any hints appreciated.
(Removed wrong proof)