For a first-order language with equality, no functions, and no relations, suppose we have a formula in this language $\alpha$ with a single free variable $x$. Consider a structure $\mathcal{A}$ in our language and consider a set $D = \{t \in |\mathcal{A}|: \mathcal{A} \vDash \alpha [s(x|t)]\}$ where $s(x|t)$ is an assignment function fixing $x$ to be $t$.
I would like to show that $D$ is either the entire domain of $\mathcal{A}$, which we denote $|\mathcal{A}|$ or $\emptyset$.
Here is what I have: I'm trying to inductively build my formula $\varphi$ and am not sure what all of the base and inductive cases are. Here's what I have so far (recall that $\varphi$ must have one and only one free variable):
Base case a) $\varphi$ is $x=x$; can be shown by seeing that for all assignment functions, where each assignment function assigns $x$ to a different element of the domain of $\mathcal{A}$, we have $a=a$. Thus $D = |\mathcal{A}|$
Base case b) $\forall x ( x=y)$; not going to go into how to show this base case, but I know how
Base case c) $\exists x (x = y)$; same deal that I wont go into how to show this base case here, but I know how
Assuming the hypothesis that $D= $ either $|\mathcal{A}|$ or $\emptyset$:
Inductive step 1:$\varphi_1 \rightarrow \varphi_2$ where $\varphi_1$ and $\varphi_2$ have the same free variable. Not going to go into details, but know how to show this
Ind step 2: $\neg \varphi$ (not going into details)
So I think I'm missing a couple of base cases for the quantifiers and would like to know what they are. Also are the rest of my cases right?
Your approach is pretty much hopeless as written. The problem is that intermediate steps of the induction may involve additional free variables, which later become bound as you add more quantifiers. For instance, consider a formula like $$\exists z\forall y(x=y\vee\exists w(z=w\wedge \neg x=w)).$$ You definitely can't build a formula like that using just the base cases and induction steps you've mentioned. Indeed, there is no way at all to build this formula step by step in terms of formulas with only one free variable; it is essential that we can use subformulas like $z=w\wedge \neg x=w$ which have many free variables.
To put it another way, formulas are built up inductively, but formulas with one free variable are NOT built up inductively. The only way to get an inductive proof about formulas with one free variable is to first prove something about all formulas and then just consider the special case where there is only one free variable.
This can be done to solve this problem, but it is considerably more complicated, since you have to first generalize the theorem statement to be valid for arbitrary formulas. Here is what turns out to be the correct generalization: given a formula $\varphi(x_1,\dots,x_n)$ with free variables $x_1,\dots,x_n$, the truth of $\varphi(a_1,\dots,a_n)$ for $a_1,\dots,a_n\in|\mathcal{A}|$ depends only on the set $\{(i,j):a_i=a_j\}$. That is, if $a_1,\dots,a_n,b_1,\dots,b_n\in|\mathcal{A}|$ and $\{(i,j):a_i=a_j\}=\{(i,j):b_i=b_j\}$, then $\mathcal{A}\vDash\varphi(a_1,\dots,a_n)$ iff $\mathcal{A}\vDash\varphi(b_1,\dots,b_n)$. This statement is now straightforward (if a bit messy) to prove by induction on formulas.
In the particular case that $n=1$, the set $\{(i,j):a_i=a_j\}$ can only be $\{(1,1)\}$, and so we conclude that $\mathcal{A}\vDash\varphi(a_1)$ iff $\mathcal{A}\vDash\varphi(b_1)$ for any $a_1,b_1\in|\mathcal{A}|$. In other words, $\varphi$ is either always true or never true, so $D=|\mathcal{A}|$ or $D=\emptyset$.
The moral here is that it is much easier to use automorphisms as explained in Rene Schipperus's answer than to try to work directly with formulas by induction.