prove that infinite vector spaces are elementarily equivalent let F be a field. I want to show that any two infinite F-vector spaces are elementarily equivalent using model theory.
Setup: Let $L$ be first order language of vector spaces over F. Non-logical symbols of $L$ are a constant $0$, a binary function symbol $+$, and a unary function symbol $F_a$ for any $a ∈ F$. For an $F$-vector space $V$ , we regard $V$ as an $L$-structure in the obvious way: 0 is interpreted by the identity element of $V$ , + is interpreted by the addition of $V$ , and each $F_a$ is interpreted by the operation of scalar multiplication by $a$.
My attempt is to take infinite F-vector spaces $A \subseteq B$, and show that A is the elementary substructure of $B$, i.e. $A$≺$B$. For this I would have to show that for every finite sequence $a_1,...,a_m \in A$ and $b\in B$, there exists an automorphism of $B$ that fixes the $a_1,...,a_m$ and sends the $b$ into $A$. How is this done rigorously?
 A: You want to show that the theory $T$ of infinite $F$-vector spaces is complete. The easiest way to do this is by the Łoś-Vaught test. For any cardinal $\kappa>|F|$, $T$ is $\kappa$-categorical. Since $T$ also has no finite models, the Łoś-Vaught test tells us that $T$ is complete.
The proof of the Łoś-Vaught test and $\kappa$-categoricity of $T$ are both so easy that we can reformulate this argument as a short direct proof:
Suppose $V$ and $W$ are two infinite $F$-vector spaces. Let $\kappa>|T|$ be a cardinal. By Löwenheim-Skolem, there are $F$-vector spaces $V'$ and $W'$ of cardinality $\kappa$ such that $V\equiv V'$ and $W \equiv W'$. Pick a basis $B_{V'}$ for $V'$ and a basis $B_{W'}$ for $W'$. Since $|V| = |W| = \kappa$ and $|F|<\kappa$, by counting we must have $|B_{V'}| = |B_{W'}| = \kappa$. Any bijection $B_{V'}\to B_{W'}$ extends to an isomorphism $V'\cong W'$. Thus $V\equiv V'\equiv W'\equiv W$ so $V$ and $W$ are elementarily equivalent.

Now what about your attempted proof? You want to show that if $A\subseteq B$ are models of $T$, then $A\preceq B$. Showing this would actually establish that $T$ is model complete. Despite the name, this is a totally different condition: there are theories that are complete but not model complete (e.g. the complete theory of $(\mathbb{N},<)$) and theories that are model complete but not complete (e.g. the theory of algebraically closed fields of arbitrary characteristic).
However, if you show that $T$ is model complete (which it is) and additionally that there is a model $M\models T$ that embeds in any other model (you can use an $F$-vector space of dimension $1$ if $F$ is infinite or an $F$-vector space of dimension $\aleph_0$ if $F$ is finite), then you can conclude that $T$ is complete. This is because for any models $V$ and $W$, there are embeddings $f\colon M\to V$ and $g\colon M\to W$, and by model-completeness $f$ and $g$ are elementary embeddings. So $V\equiv M\equiv W$.
Ok, so let's try to follow your strategy to show that $T$ is model complete. We have infinite $F$-vector spaces $A\subseteq B$, and you want to show that $A\preceq B$ by applying the Tarski-Vaught test. It suffices to show that for every finite sequence $a_1,\dots,a_m\in A$ and $b\in B$, there exists an automorphism of $B$ that fixes the $a_1,\dots,a_m$ and maps $b$ into $A$.
If $A$ is infinite-dimensional (which it must be if $F$ is a finite field), we can do this. If $b\in A$ already, we're done, using the identity automorphism. So assume $b\notin A$. First pick a basis $B_0$ for the subspace $\text{Span}(\{a_1,\dots,a_m\})\subseteq A$. Extend $B_0$ to a basis $B_1$ for $A$. Since $b\notin A$, $B_2 = B_1\cup \{b\}$ is linearly independent. Extend $B_2$ to a basis for $B_3$ for $B$. Now since $B_0$ is finite and $A$ is infinite-dimensional, we can pick some $b'\in B_1\setminus B_0$. Let $\sigma\colon B_3\to B_3$ be any bijection which fixes $B_0$ and maps $b\mapsto b'$. Then $\sigma$ extends to an automorphism of $B$ which fixes $\text{Span}(\{a_1,\dots,a_m\})$ and maps $b$ to $b'\in B_1\subseteq A$.
Unfortunately, there's no hope of doing this if $A$ is finite-dimensional (which it might be, when $F$ is an infinite field). Consider the case when $a_1,\dots,a_m$ is a basis for $A$. Any automorphism of $B$ which fixes $a_1,\dots,a_m$ fixes all of $A$, so it cannot map an element of $B\setminus A$ into $A$.
A better approach to proving model-completeness is to prove directly that $T$ has quantifier elimination.
