Let $A_n$ be a sequence of bounded, self-adjoint operators converging to $A$ in the strong operator topology. Even if the spectra of all $A_n$ coincide, the spectrum of $A$ could be a strict subset of these spectra (as noted here).
I am interested in the reverse direction: If $\forall n:\sigma(A_n)\subseteq\sigma(A_{n+1})$ (or even just with equality) , do we have $\sigma(A)\subseteq \bigcup_n\sigma(A_n)$ or could the spectrum of the limit contain more values?
You need to take the closure on the right, but then it is true:
Let $\lambda \in \mathbb{R} \setminus \overline{\bigcup\limits_{n \in \mathbb{N}} \sigma(A_n)}$. Then there is an $\varepsilon > 0$ such that the open ball $B(\lambda,\varepsilon)$ does not intersect with any $\sigma(A_n)$. In particular, as $A_n - \lambda I$ is self-adjoint, $\inf\limits_{\|x\| = 1} \|(A_n - \lambda I)x\| \geq \varepsilon$. It follows that $$\|(A - \lambda I)x\| = \lim\limits_{n \to \infty} \|(A_n - \lambda I)x\| \geq \varepsilon\|x\|$$ for all $x \in H$. As $A - \lambda I$ is self-adjoint, this implies that $A - \lambda I$ is invertible. Thus $\lambda \in \mathbb{R} \setminus \sigma(A)$.