Let's say I have a scalar product : $v$ on $C^0(\mathbb{R}, \mathbb{R})$, then by Cauchy-Scwartz inequality I have :
$$\forall f,g \in(C^0(\mathbb{R}, \mathbb{R}),v)\quad v(f,g) \leq \| f\|_v \|g\|_v$$
Now let's say I have an other scalar product : $w$ on $C^0(\mathbb{R}, \mathbb{R})$, then do I have $\forall f, g \in (C^0(\mathbb{R}, \mathbb{R}),\color{red}{v})$ :
$$ w(f,g) \leq \| f\|_w \|g\|_w$$
(so is this inequality true, even-if $f,g$ "lives" in a different inner-product space)
It seems like you are confused by notation. Suppose $X$ is a real vector space and $v,w$ are two inner products on $X;$ that is $(X,v)$ and $(X,w)$ are inner product spaces.
The notation "let $x \in (X,v)$" just says "let $x \in X$." Sometimes it's convenient to remind the reader that we are considering $X$ as an inner product space (with inner product $v$), but technically it is not giving any additional information. Hence the spaces $(X,v)$ and $(X,w)$ may not be "the same" as inner product spaces, but any $x \in X$ "lives" in both spaces nevertheless.
As a result the Cauchy-Schwarz inequality holds for any $x,y \in X,$ with respect to any choice of scalar product on $X.$