The hyperplane separation theorem reads
Let $A$ and $B$ be two disjoint nonempty convex subsets of $\mathbb{R}^n$. Then there exist a nonzero vector $v$ and a real number $c$ such that
$$ \langle x,v\rangle \geq c,{\text{ and }}\langle y,v\rangle \leq c $$ for all $x \in A$ and $y \in B$.
I want to show that if $A$ is open then we actually have strong separation, i.e. $ \langle x,v\rangle > c$ for all $x \in A$.
Suppose $\langle x,v\rangle =\sum_{i=1}^n x_i v_i= c$ for some $x\in A$. Since $A$ is open, there exist $\epsilon>0$ such that $B(x,\epsilon)\subset A$. Since $v\neq 0$ there exist some $j$ such that $v_j\neq 0$. If $v_j>0$, then we can find $\delta>0$ small enough so that replacing $x_j$ with $x_j-\delta$ in $x$ results in a vector $x' \in A$ with $\langle x',v\rangle <c$. Similarly, if $v_j<0$, then we can find $\delta>0$ small enough so that replacing $x_j$ with $x_j+\delta$ in $x$ results in a vector $x' \in A$ with $\langle x',v\rangle <c$. In either case this contradicts the separating hyperplane property, and so we have strong separation.
Is this correct?