Let $\mathbf{B}$ be the algebra $\mathbf{A}\times\mathbf{A}$ together with the new fundamental operations listed in the proof in Burris & Sankappanavar:
$\begin{align*} \langle a,a\rangle&\qquad\text{for }a\in A\\ s\big(\langle a,b\rangle\big)=\langle b,a\rangle&\\ t\big(\langle a,b\rangle,\langle c,d\rangle\big)=\begin{cases}\langle a,d\rangle\\ \langle a,b\rangle \end{cases}&\qquad\begin{array}{l}\text{if }b=c\\\text{otherwise}\end{array}\\ e_\sigma\big(\langle a,b\rangle\big)=\langle\sigma a,\sigma b\rangle&\qquad\text{for }\sigma\text{ and endomorphism of }\mathbf{A}\;. \end{align*}$
The key step is the assertion that $\theta$ is a fully invariant congruence on $\mathbf{A}$ iff $\theta$ is a subuniverse of of $\mathbf{B}$.
Let $\theta$ be a congruence on $\mathbf{A}$, and note that $\theta\subseteq A\times A$. If $\theta$ is a subuniverse of $\mathbb{B}$, then by definition $\theta$ is closed under the fundamental operations of $\mathbf{B}$. In particular, $\theta$ is closed under $e_\sigma$ for each endomorphism $\sigma$ of $\mathbf{A}$, which is exactly what it means for $\theta$ to be fully invariant.
Now suppose that $\theta$ is fully invariant. By the definition of congruence (Definition 5.1) $\theta$ is an equivalence relation, so it’s closed under the first three operations listed above. Definition 5.1 also requires that $\theta$ be closed under the fundamental operations of $\mathbf{A}$. Finally, $\theta$ is closed under the new operations $e_\sigma$ because by hypothesis it’s fully invariant. $\dashv$
It follows from Theorem 3.2 that $\Theta_{\text{FI}}$ is an algebraic closure operator: for any $S\subseteq A\times A$,
$\begin{align*} \Theta_{\text{FI}}(S)&=\bigcap\{\theta\in\operatorname{Con}_{\text{FI}}(\mathbf{A}):S\subseteq\theta\}\\ &=\bigcap\{B\subseteq A\times A:S\subseteq B\text{ and }B\text{ is a subuniverse of }\mathbf{B}\}\\ &=\operatorname{Sg}(S)\;. \end{align*}$