Let $\mathcal A\subset\mathcal B(E)$ be a $C^*$ sub-algebra inside the algebra of bounded operators in a Hilbert space $E$,with $I\in \mathcal A$. For $T\in \mathcal A$,let $\sigma(T)$ denote the spectrum (as an element of $\mathcal B(E)$) and $\sigma_{\mathcal A}(T)$ denote the spectrum as an element of $\mathcal A$,so that $\sigma(T)\subset \sigma_{\mathcal A}(T).$
We have to show the following three things:
$1$. boundary of $\sigma_{\mathcal A}(T)=$ boundary of $\sigma(T)$.
$2$. If $T=T^*$, then $\sigma_{\mathcal A}(T)=\sigma(T)$. This part easily follows from part $1$ and the fact that $\sigma(T)\subset \mathbb R$.
$3$. $\sigma_{\mathcal A}(T)=\sigma(T)$ for all $T\in \mathcal A$. This also follows easily because if $S=T-\lambda I$ is invertible in $\mathcal B(E)$,then $S^{-1}=S^*(SS^*)^{-1}\in \mathcal A$ since $SS^*$ is self-adjoint.(Using part $2$.)
So, I am only stuck with part $1$.
These kind of topics are very new to me. Any help or hint would be appreciated. Thanks in advance.
I do not think you actually need to show that $$ \partial \sigma_{\mathcal A}(T) = \partial \sigma(T) . $$ It would suffice to show that $$\partial \sigma_{\mathcal A}(T) \subseteq \sigma(T) \tag{$*$} $$ Then if $T=T^*$, you get that $\partial \sigma _{\mathcal A} (T) \subseteq \sigma(T) \subseteq \mathbb R$ and so $$\sigma _{\mathcal A} (T) = \partial \sigma _{\mathcal A} (T) \subseteq \sigma(T) \subseteq \sigma_{\mathcal A}(T).$$
To prove $(*)$, let $λ \in \partial \sigma_{\mathcal A}(T) .$ Pick a sequence $λ_n \notin \sigma_{\mathcal A}(T) $ such that $λ_n \to λ$, so that $(T-\lambda_n)^{-1} \in \mathcal A$. If it were the case that $\lambda \notin \sigma(T)$ then $(T-λ)^{-1} \in B(H)$, and since inversion is continuous, $$(T-λ)^{-1} = \lim_n (T-λ_n)^{-1} \in \overline{\mathcal A} = \mathcal A .$$ So $λ \notin \sigma_{\mathcal A}(T)$, which is impossible since $λ\in \partial \sigma_{\mathcal A}(T) $.
$\textbf{Edit:}$ Since $\sigma_{\mathcal A}(T)$ is a subset of the real line, each of its points must be a boundary point. That is, $\sigma _{\mathcal A} (T) = \partial \sigma _{\mathcal A} (T) $. This is due to the fact that any subset of the real line must have emtpy interior in $\mathbb C$. [Recall that $\partial E= \overline{E} \setminus \operatorname{int} E$.]