Baire category theorem in a Banach space

Question

For any two distinct $u,v$ in a countable dense subset of separable real Banach space $X$, let $S(u,v) = \{f \in Y \mid f(u)=f(v)\}$, where $Y$ is the dual space of $X$. Each of $S(u,v)$ is a proper closed subspace of $Y$. How can I prove that the Baire category theorem implies that there exists $f$ in $Y$ which is not in any of countably many sub-spaces $S(u,v)$?

Answer 1

Outline: prove that each $S$ is closed with empty interior; then the Baire category theorem implies that their union has empty interior; but $Y$ itself does not have empty interior, so their union is a proper subset of $Y$.

So the problem reduces to showing that $S(u,v)$ is closed and has empty interior. Note that $S(u,v)$ is the kernel of the map $i : Y \to \mathbb{R}$ defined by $i(f)=f(u-v)$. It remains to show that:

1. $i$ is continuous and linear
2. $S$ is a closed proper subspace of $Y$
3. any proper subspace of a Banach space has empty interior.

I'll offer additional hints on these steps upon request.