A set $\Delta$ of Formulae is deductively closed iff $\Delta$ $\vdash$ $\sigma$ implies that $\sigma$ $\in$ $\Delta$.
$\Gamma$ is a set of sentences.
Define $\Sigma_{\mathcal{A}}$ as all sentences in the language which are true in the structure $\mathcal{A}$. Show that $\Delta (\Gamma)$ (that is, the deductive closure of $\Gamma$) is the intersection of all $\Sigma_{\mathcal{A}}$ so that all sentences in $\Gamma$ are true in the structure $\mathcal{A}$.
Now my question is less about how to show this and more about whether the above even makes sense. For any Structure, $\mathcal{A}$, shouldn't there be only one $\mathcal{A}$? How then can we form an intersection? Don't we call the $\Sigma_{\mathcal{A}}$ the complete theory of $\mathcal{A}$, or $Th(\mathcal{A})$? I would think that you could have multiple $\Sigma_{\mathcal{A}}$ in $\mathcal{A}$ if $\Sigma_{\mathcal{A}}$ was a set of formulas with free variables, such that different interpretations of variables would yield different sets, but sentences should be true regardless of how free variables map in the domain of $\mathcal{A}$. Any hints much appreciated :)
Yes, from your definition, it looks like $\Sigma_{\mathcal A}$ is just an alternative notation for $\operatorname{Th}(\mathcal A)$.
The claim you're supposed to prove must be $$ \Delta(\Gamma) = \bigcap_{\mathcal A\text{ such that }\mathcal A \vDash \Gamma} \Sigma_{\mathcal A} $$ so $\mathcal A$ is not a particular structure; instead you're taking the intersection over all structures that satisfy $\Gamma$. (And if no structure satisfies $\Gamma$, then you're implicitly supposed to take the intersection to be the set of all sentences in your language).
The above claim is equivalent to the Soundness and Completeness theorems for your deductive system; I suspect you already know these theorems and just need to show they can be phrased in the above way. (In fact, the $\subseteq$ direction is soundness, and $\supseteq$ is completeness).