In chapter 1 of Milne's Étale cohomology book from 1980, Theorem 3.14 states that :
If $f: Y\longrightarrow X$ is étale in some open neighbourhood of a point $y\in Y$, then there are affine open neighbourhoods $V$ and $Y$ of $y$ and $f(y)$ respectively such that $f$ restricted to $V$ is a standard étale morphism.
The proof begins with the assumption that we may restrict to the case $Y=Spec C$ and $X=Spec A$ where $C$ is a finitely generated $A$ algebra. I take this to be a consequence of the fact that the map is unramified when restricted to the neighbourhood of $y$ on which $f$ is étale, meaning that the map is locally of finite type there.
It is then claimed that we need to show that for some $c\not\in \mathfrak{q}$ where $\mathfrak{q}$ is the ideal corresponding to $y$, we have $C_c$ is isomorphic as an $A$-algebra to a standard étale algebra i.e, something of the form $[A[t]/(P(t))]_b$ where $P(t)$ is monic and $P'$ is invertible in $[A[t]/(P(t))]_b$. So if I interpret this correctly, we are finding a basic Zariski open neighbourhood containing $y$ as required where the map is standard étale.
Then it is claimed that "because everything is finite over $A$, it suffices to do this with $A$ replaced by $A_{\mathfrak{p}}$", where $\mathfrak{p}$ is the ideal corresponding to $f(y)$. I don't understand why this is the case.
You missed the key point that we can need only consider the case where we have a homomorphism $A\rightarrow C$ with $C$ a finite (not merely finitely generated) $A$-algebra. This uses the affine form of Zariski's main theorem. Now suppose that we have a polynomial $P(t)$ of the correct form such that the $A_{\mathfrak{p}{}}$-algebra $C\otimes_{A}A_{\mathfrak{p}{}}$ is isomorphic to $A_{\mathfrak{p}{}}[t]/(P(t))$. After replacing $A$ with $A_{b}$ we may suppose that $P(t)$ has coefficients in $A$. Now we have two finite $A$-algebra $A[t]/(P(t))$ and $C$. The set where they are isomorphic is open in $Spec(A)$ and nonempty.