Q:

Why is compactness in logic called compactness?

In logic, a semantics is said to be compact iff if every finite subset of a set of sentences has a model, then so to does the entire set. 
Most logic texts either don't explain the terminology, or allude to the topological property of compactness. I see an analogy as, given a topological space X and a subset of it S, S is compact iff for every open cover of S, there is a finite subcover of S. But, it doesn't seem strong enough to justify the terminology. 
Is there more to the choice of the terminology in logic than this analogy?

A:

The Compactness Theorem is equivalent to the compactness of the Stone space of the Lindenbaum–Tarski algebra of the first-order language $L$. (This is also the space of $0$-types over the empty theory.)
A point in the Stone space $S_L$ is a complete theory $T$ in the language $L$. That is, $T$ is a set of sentences of $L$ which is closed under logical deduction and contains exactly one of $\sigma$ or $\lnot\sigma$ for every sentence $\sigma$ of the language. The topology on the set of types has for basis the open sets $U(\sigma) = \{T:\sigma\in T\}$ for every sentence $\sigma$ of $L$. Note that these are all clopen sets since $U(\lnot\sigma)$ is complementary to $U(\sigma)$.
To see how the Compactness Theorem implies the compactness of $S_L$, suppose the basic open sets $U(\sigma_i)$, $i\in I$, form a cover of $S_L$. This means that every complete theory $T$ contains at least one of the sentences $\sigma_i$. I claim that this cover has a finite subcover. If not, then the set $\{\lnot\sigma_i:i\in I\}$ is finitely consistent. By the Compactness Theorem, the set consistent and hence (by Zorn's Lemma) is contained in a maximally consistent set $T$. This theory $T$ is a point of the Stone space which is not contained in any $U(\sigma_i)$, which contradicts our hypothesis that the $U(\sigma_i)$, $i\in I$, form a cover of the space.
To see how the compactness of $S_L$ implies the Compactness Theorem, suppose that $\{\sigma_i:i\in I\}$ is an inconsistent set of sentences in $L$. Then $U(\lnot\sigma_i),i\in I$ forms a cover of $S_L$. This cover has a finite subcover, which corresponds to a finite inconsistent subset of $\{\sigma_i:i\in I\}$. Therefore, every inconsistent set has a finite inconsistent subset, which is the contrapositive of the Compactness Theorem.

A:

The analogy for the compactness theorem for propositional calculus is as follows. Let $p_i $ be propositional variables; together, they take values in the product space $2^{\mathbb{N}}$.  Suppose we have a collection of statements $S_t$ in these boolean variables such that every finite subset is satisfiable. Then I claim that we can prove that they are all simultaneously satisfiable by using a compactness argument.
Let $F$ be a finite set. Then the set of all truth assignments (this is a subset of $2^{\mathbb{N}}$) which satisfy $S_t$ for $t \in F$ is a closed set $V_F$ of assignments satisfying the sentences in $F$.  The intersection of any finitely many of the $V_F$ is nonempty, so by the finite intersection property, the intersection of all of them is nonempty (since the product space is compact), whence any truth in this intersection satisfies all the statements.
I don't know how this works in predicate logic.

A:

Lemma: A topological space $X$ is compact if and only if for every collection $\mathcal{C}$ of closed sets with the finite intersection property has nonempty intersection over the collection.
Proposition: $\mathbb{M}(\mathcal{L})$ is compact if and only if every finitely satisfiable $\mathcal{L}$-theory is satisfiable.
Proof: Consider the space $\mathbb{M}(\mathcal{L})=\{\Phi \; | \; \Phi \; \text{is a maximal} \; \mathcal{L} \text{-theory}\}$. For each $\mathcal{L}$-sentence $\varphi$ let $[(\varphi)]=\{ \Phi \in \mathbb{M}(\mathcal{L}) \; | \; \varphi \in \Phi\}$. A subbase $\cal{B}$ for a topology on $\mathbb{M}(\mathcal{L})$ is given by the sets $[(\varphi)]$. That is, the open subsets of $\mathbb{M}(\mathcal{L})$ are the union of the finite intersections of elements of $\mathcal{B}$. To see that $\mathcal{B}$ defines a topology on note that $[(\forall x(x\neq x))]=\emptyset$ and $[(\forall x(x=x))]=\mathbb{M}(\mathcal{L})$. Furthermore, arbitrary unions are already defined to be in the topology and finite intersections are in the topology since $[(\varphi)]\cap [(\theta)]=[(\varphi \wedge \theta)]$, which is defined to be open. 
Assume logical compactness. Note that $\bigcap_{i\in I} [(\varphi_i)]=\emptyset$ if and only if it is not satisfiable. Let $\mathcal{C}$ be a subcollection of the subbasis of the topology of $\mathbb{M}(\mathcal{L})$ with the finite intersection property. Then every finite subset of $\mathcal{C}$ is satisfiable and by the compactness theorem this implies $\bigcap [(\varphi)]$ which ranges over all the elements of $\cal{C}$ is satisfiable, hence
$$
\bigcap_{[(\varphi)] \in \cal{B}} [(\varphi)]\neq \emptyset.
$$
Thus $\mathbb{M}(\mathcal{L})$ is compact. 
Assume $\mathbb{M}(\mathcal{L})$ is compact. Let $\Phi$ be an $\mathcal{L}$-theory in which is finitely satisfiable. Let $\mathcal{C}_\Phi=\{[(\varphi)] \; | \; \varphi \in \Phi\}$. Every element of $\mathcal{C}_\Phi$ is closed and every finite subset of $\mathcal{C}_\Phi$ has nonempty intersection since $\Phi$ is finitely satisfiable. Since $\mathbb{M}(\mathcal{L})$ is compact, $\bigcap_{\varphi \in \Phi} [(\varphi)]\neq\emptyset$, hence it is satisfiable. $\square$

