Separation holds in $L$ In Chapter 13 of Jech's Set Theory, he proves that $L$ is a model of ZF. For Separation, he writes the following:

Let $\varphi$ be a formula. Given $X,p\in L$, we wish to show that the set $Y=\{u\in X:\varphi^L(u,p)\}$ is in $L$. By the Reflection Principle (applied to the cumulative hierarchy $L_\alpha$, cf. Exercise 12.6), there exists an $\alpha$ such that $X,p\in L_\alpha$ and $Y=\{u\in X:\varphi^{L_\alpha}(u,p)\}$. Thus $Y=\{u\in L_\alpha:L_\alpha\models u\in X\land \varphi(u,p)\}$ and so $Y\in L$.

I'm not quite sure what he means by the Reflection principle applied to the cumulative hierarchy $L_\alpha$ (Exercise 12.6 has nothing to do with $L$ or the reflection principle). So far as I understand, (a strengthening of) the Reflection Principle gives that for any $\varphi$ and $M_0$, there is some $\alpha$ such that $V_{\alpha}\supseteq M_0$ reflects $\varphi$. In the case of $L$, would that mean there is some $\alpha$ with $\varphi^{L_\alpha}\leftrightarrow \varphi^L$?.
As for how this applies, my best guess is that if $X,p\in L$, then there is some $\beta$ with $X,p\in L_\beta$. We then apply this modified reflection formula to get some $\alpha$ with $\varphi^{L_\alpha}\leftrightarrow \varphi^L$, and thus $Y=\{u\in X:\varphi^{L_\alpha}(u,p)\}$. We then use the formula "$L_\alpha\models u\in X\land \varphi(u,p)$" as a witness to the definability of $Y$. Is this a correct understanding of what is going on?
 A: Yes, your understanding is largely correct. Let me state a version of reflection theorem that is used in the reasoning, as in Kunen IV 7.5:
Reflection theorem. Suppose $Z$ is a class and for each $\alpha$, $Z_\alpha$ is a set, and assume
(i) $\alpha<\beta\to Z_\alpha\subset Z_\beta$,
(ii) if $\gamma$ is a limit ordinal, $Z_\gamma=\bigcup_{\alpha<\gamma}Z_\alpha$,
(iii) $Z=\bigcup_{\alpha\in \mathrm{On}}Z_\alpha$. 
Then for any formulas $\phi_1,\ldots,\phi_n$, 
$\forall\alpha(\exists\beta>\alpha)\bigwedge_{i=1}^n(\phi_i^{Z_\beta}\leftrightarrow \phi_i^Z)$.
We can apply the theorem by letting $Z$ be $L$, and $Z_\alpha$ be $L_\alpha$.
Given $X,p\in L$, there is $\beta$ such that $X,p\in L_\beta$. Then we apply the theorem to find some $\alpha>\beta$ such that $\phi^{L_\alpha}\leftrightarrow \phi^L$. Since $\alpha>\beta$, also $X,p\in L_\alpha$ (and $X\subset L_\alpha$). Then $Y=\{ u\in X\mid \phi^{L_\alpha}(u,p)\}=\{ u\in L_\alpha\mid L_\alpha\models u\in X\wedge \phi(u,p)\}$, showing $Y\in \mathrm{def}(L_\alpha)$.
