Gödel's completeness theorem proves that in first order logic, if a theory is consistent (we cannot derive a contradiction), then it has a model.
As discussed in this question, there are theories which are consistent but do not have a model.
In particular, the completeness theorem does not apply to second order logic. I would like to understand better: what is it in second order logic that make the usual proofs of completeness (of first order logic) fail?
Noah Schweber mentions in a comment "an explanation of why the proofs of completeness for FOL inevitably break down for SOL". I will start with something more modest: an explanation of how the usual Henkin-style proof of the completeness theorem fails for SOL.
Just to fix ideas (you probably know all this), second-order logic is a two-sorted affair, with its universe consisting of elements and sets of elements. Lower-case variables range over the elements, and upper case over the sets. Quantification over sets are allowed, so we have both $\forall x$ and $\forall X$. A second-order model has an element-domain $U$ and the set-domain $P(U)$, the "true" power-set of $U$. The upper-case variables range over $P(U)$. That last point is crucial.
"Viewed from a thousand feet", the Henkin proof of the FOL completeness theorem expands the consistent theory $T$ to a complete theory $T^*$ with a bunch of new constants (countably many, if $T$ is countable). $T^*$ is deductively closed. The proof constructs a model $M$ whose elements are the new constants. Then the proof shows that for any sentence $\varphi$, $\varphi\in T^*$ if and only if $M$ satisfies $\varphi$. This is by induction on complexity. (You also need to mod out by the = relation if you want a model in which = really means "equals", but we can ignore this detail.)
In particular, if we have an existential statement $\exists x\varphi(x)$, the construction gives you a constant $c_\varphi$ such that $\exists x\varphi(x)\in T^*$ iff $\varphi(c_\varphi)\in T^*$.
The crucial induction argument for existential statements goes like this: if $\exists x\varphi(x)\in T^*$, then $\varphi(c_\varphi)\in T^*$, so by inductive hypothesis, $M\models \varphi(c_\varphi)$. But then $M\models \exists\varphi(x)$. Conversely, if $M\models \exists\varphi(x)$, then $M\models \varphi(c)$ for some $c$ of the model, so by inductive hypothesis $\varphi(c)\in T^*$ and so $\exists x\varphi(x)\in T^*$ (because $T^*$ is deductively closed).
We can mimic this construction for a second-order theory. Our expanded theory $T^*$ will have a bunch of element-constants for elements, and a bunch of set-constants for sets of elements. The model will be built out of these constants. In particular, the sets of the model will correspond to the set-constants. But note: except in very special cases, there will be many sets of elements in $P(U)$ that do not correspond to any of the set-constants.
In other words, the model we construct has a universe consisting of an element-domain $U$, and a set-domain $V\subseteq P(U)$. Usually $V$ is a proper subset of $P(U)$. The crucial induction step, sketched above, shows only that $\exists X\varphi(X)\in T^*$ iff there is a set-constant $C$ such that $M\models\varphi(C)$.
That means we can have $M\models\exists X\varphi(X)$ while $\exists X\varphi(X)\not\in T^*$, because the set $\bar{X}$ with $M\models\varphi(\bar{X})$ doesn't correspond to any of the set-constants. Turned around for universal statements, if $\forall X\varphi(X)\in T^*$, then $\varphi(\bar{X})$ holds in $M$ for all "named" sets, but not necessarily for all sets.
In essence, although we've been pretending to deal with SOL, really we've been treating the two-sorted theory as a first-order theory. The quantifier $\forall X$ isn't interpreted as meaning "for all subsets of the universe of elements", but just "for all the sets named by the set-constants".
Here's an example, similar to the example in the linked answer. Let $T_0$ be the second-order theory with relation symbols $<$ and $\in$ and postulates
The second bullet is the only second order postulate: $$\forall S(\exists x\,x\in S\to \exists x(x\in S\wedge \forall y(y\in S\to\neg y<x))$$ which I will express more concisely as $$\forall S(S\neq\varnothing\to \text{has-min}(S))$$ These postulates insure that the universe is order-equivalent to $\mathbb{N}$. Now add constants $\{a_i|i=0,1,2,\ldots\}$, postulates $a_i<a_j$ whenever $i<j$ and $\neg a_i<a_j$ whenever $\neg i<j$, and a constant $c$ with postulates $\{a_i<c|i=0,1,\ldots\}$. Call the resulting theory $T$.
The Henkin-style proof produces a model with an "infinite" element $c$. Using the immediate predecessor postulate, it follows that there is a set $S=\{c, c-1, c-2,\ldots\}$. But the Henkin construction never "names" this set. $S$ belongs to the power-set of the element-domain, but not to the set-domain $V$.
As for the "inevitable breakdown" of proofs of the "SOL completeness theorem": the only conclusive argument for this is the existence of counter-examples to the proposed theorem. Suppose we had a dozen different attempted proofs. Suppose they all broke down in much that same way. Perhaps there could be a 13th approach that steered around the common obstacle.
That said, the issue I've explained for the attempted Henkin-style proof also appears, in much the same way, in the other proofs I know for the FOL completeness theorem.
