What goes wrong when you try to reflect infinitely many formulas? The reflection principle in ZFC shows that you can construct a set that reflects finitely many formulas. Suppose we wanted to reflect {$\phi_n$} and we construct a set $M_n$ to reflect $\phi_1, ... , \phi_n$ with $M_n \subset M_{n+1}$. Why doesn't $M=\cup M_n$ reflect all of the {$\phi_n$}? I know this has to be false since we shouldn't be able to reflect all of ZFC with a set, but when we try to prove that this works, it seems that the existential step goes through since those formulas can only have finitely many variables, and when you pick finitely many elements of $M$, they will all be in some $M_k$. (By the "existential step" I mean proving that : if $\exists x\phi_i(u_1, ..., u_j,x)$ then $(\exists x \in M)\phi_i(u_1,...,u_j,x)$ )
Relatedly, using the reflection principle, we can show that ZFC proves that any finite set of axioms from ZFC are consistent. Since ZFC can also prove compactness, why doesn't this show that ZFC proves its own consistency, which contradicts Gödel's theorem?
 A: Let me first address your concrete argument. The reflection theorem says that for any finite collection of formulas ZFC proves that there is a set reflecting those formulas. Note that this does not say that ZFC proves that for any finite collection of formulas there is a set reflecting those formulas. This is a crucial distinction. The latter would be a single assertion, while the former is a collection of assertions, one for each collection of formulas. All of these separate statements live their lives pretty much independently of one another.
Now look at a given model $X$ of ZFC and your collection of formulas $\varphi_1,\varphi_2,\dots$ Applying the reflection theorem infinitely often (one instance per each finite subcollection of the $\varphi_i$) we indeed get your $M_n\in X$ which reflect their relevant formulas. But $X$ will not in general have the function $n\mapsto M_n$ precisely because the reflection theorem is not a single assertion (with parameter $n$) to be used in $X$ but a metatheoretical collection of statements. And since $X$ doesn't have that sequence, your $M$ which is its union need not be an element of $X$.
Furthermore, even if we had $M\in X$ there are still issues with saying that $M$ reflects all of the $\varphi_i$.
What does it mean for a set $M$ to reflect a formula $\varphi$? It means that
$\varphi\iff \varphi^M$ where $\varphi^M$ is the relativization of $\varphi$ to $M$. Saying that there is such an $M$ amounts to saying $\exists M\colon \varphi\iff\varphi^M$. Similarly, saying that there is an $M$ which reflects finitely many formulas $\varphi_1,\dots,\varphi_n$ is just saying 
$$\exists M\colon\bigwedge_{i<n}(\varphi_i\iff\varphi_i^M)$$
Now consider what it would mean to say that $M$ reflects infinitely many formulas $\varphi_1,\varphi_2,\dots$ While this statement has a clear model-theoretic meaning, we simply cannot formally express it in the language of set theory as we did before. The best we can do is again to use a metatheoretic collection of statements, one for each $\varphi_i$, saying that $M$ reflects $\varphi_i$. 
