Suppose A infinite L structure, such that $\phi(\overline{x})$ is a formula such that $\phi(A)$ - the set defined by $\phi$ is denumerable. Find B $\equiv$ A such that $\phi(B)$ uncountable.
heres my idea:
in the $\overline{x} = x$ (ie 1 tuple case), this is a standard compactness argument. augment L with uncounably many fresh constants; create $\Sigma$ = union of original theory and $\phi(c)$ for all new constants and statements saying the new constants are not equal. then use the fact that $\phi(A)$ infinite to show that any finite subset has a model. by compactness, done.
all other cases "reduce" to the 1 tuple case, for example suppose
$\phi (a_1, a_2, a_3), \phi (a_1, a_2, a_5), \phi (a_7, a_2, a_5),...$ all hold.
claim: at least one of the (set of projections of) tuple components must contain infinitely many different variables. for suppose otherwise. then each component only contains finitely many different variable, so finitely many different tuples for which $\phi$ holds. but this is a contradiction.
so we can just make the set {$\phi(\hat{a_1}, \hat{a_2}, W)$} where $\hat{k}$ denotes intoa set of fresh constants meant to interpreted as elements in A, and W ranges over the uncountable many fresh constants added. we only add in $\phi(\hat{a_1}, \hat{a_2}, W)$ if there is some corresponding tuple $\phi(a_1,a_2,a_3)$ that holds and the third compent is the column that contains infinitely many fresh variables. the argument then proceeds essentially as above.
Does this work? Is there a cleaner way to do this?
The way you expressed your argument is not so clear, so I can't say for sure if it works. But yes, there's a cleaner way.
Let's say $\varphi(\overline{x})$ has $n$ free variables $x_1,\dots,x_n$. Let $\theta_=(\overline{x},\overline{y})$ be the formula $\bigwedge_{i=1}^n x_i=y_i$. Note that $\theta_=(\overline{a},\overline{b})$ is true if and only if $\overline{a}$ and $\overline{b}$ are equal as $n$-tuples.
Now introduce uncountably many new $n$-tuples of constants: $(\overline{a}_\alpha)_{\alpha<\kappa}$ for an uncountable cardinal $\kappa$, where $\overline{a}_\alpha = (a^1_\alpha,\dots,a^n_\alpha)$. By the same compactness argument you described, the theory $$\Sigma\cup \{\varphi(\overline{a}_\alpha)\mid \alpha<\kappa\}\cup \{\lnot \theta_=(\overline{a}_\alpha,\overline{a}_\beta)\mid \alpha<\beta<\kappa\}$$ is consistent, and we're done.