It seems that the original poster's question is why the constructed model is an interpretation, i.e., closed under functional symbols:I assume that the phrase I emphasized means that the functional symbol f is in the language. Is it so? If yes, this has been shown by clic-clac. Also: (here $\displaystyle D^*$ is the same as $\displaystyle <D_0>$ above). Note that $\displaystyle D^*$ occurs in both sides, which shows that it is closed.My logic professor told me to show that the interpretation I created had an countable domain, "closed under f" (if there was one function symbolf), and such that the restriction of T would make A true.

To describe this in more detail, let's define analgebraas an interpretation with all relational symbols removed. I.e., an algebra is a domain plus interpretations of constants and functional symbolson this domain(let's from now on include constants in functional symbols). So an algebra is closed under functions by definition. Let me denote the interpretation of a functional symbol $\displaystyle f$ or a predicate $\displaystyle R$ by $\displaystyle [[f]]$ or $\displaystyle [[R]]$, respectively.

The suggested proof of Löwenheim-Skolem theorem consists of three parts. In the first, we take the algebra part $\displaystyle \langle D, [[\cdot]]^f\rangle$ of the given uncountable interpretation $\displaystyle T=\langle D, [[\cdot]]^f, [[\cdot]]^R\rangle$ and build its countable subalgebra $\displaystyle \langle D_0,[[\cdot]]_0^f\rangle$ (which is an algebra and therefore closed under functions). In the second part, we build a series of interpretations $\displaystyle T_n=\langle D_n, [[\cdot]]_n^f, [[\cdot]]_n^R\rangle$ and take its union $\displaystyle T_\omega=\langle D_\omega, [[\cdot]]_\omega^f, [[\cdot]]_\omega^R\rangle$. Finally, in the third part we show that for any formula $\displaystyle A$ in the original language, $\displaystyle T\models A$ iff $\displaystyle T_\omega\models A$. In fact, the first and second parts are pretty similar because the initial subalgebra $\displaystyle \langle D_0,[[\cdot]]_0^f\rangle$ can also be built by taking a countable union of algebras. So, the first step build a series of algebras to make them closed under functional symbols, and the second step builds a series of interpretations in order to make them closed under $\displaystyle \exists$.

The first step has been described a couple of times by clic-clac, so I am wondering if this was your question about closure. I can write it in more detail if you need.