Separation of elements of $X$ by elements of $X^*$ The following is an Exercise from Bruckner's Real Analysis:

If $x, y$ are distinct points in a normed linear space $X$, show that
there is a member of $X^*:=\mathcal{B}(X,\mathbb{R})$ that separates $x$ and $y$.

How do you approach this problem?
I don't think that this question is equivalent to "there always exists an injective bounded linear functional on $X$", because the $f \in X^*$ depends on the pair $x, y$ too such that if $x \ne y$ then $f(x) \ne f(y)$.
Possible answer? : Let $p(x)=1$ for $x \in X$, $f(x)=1$ for $x \in \operatorname{span}(x)$ and $f(x)=0$ for $x \in X - \operatorname{span}(x)$. If $x \ne y$ then either $y=ax$ for some $a \ne 1$ or $y \in X - \operatorname{span}(x)$ which in both cases $f(x) \ne f(y)$. but the problem is $f$ must be continuous because it is linear and bounded?
 A: Let $x, y \in X$, such that $x \ne y$. So, $x \ne 0$ or $y \ne 0$. Let us assume, without loss of generality, that $y \ne 0$.
Let $Y$ be the subspace generated by $y$. Note that $Y$ is a closed subspace.  We have two cases:

*

*$x \in Y$. So there is $\alpha \in \Bbb R$, such that $x= \alpha y$ and, since  $x \ne y$, we have that $\alpha \ne 1$. Now, using Hahn-Banach, we have that  there is a functional $l \in X^*$ such that $l(y) \ne 0$. It follows that $l(x) = \alpha l(y)$. Since $\alpha \ne 1$ and   $l(y) \ne 0$, we have that $l(x) \ne l(y)$.


*If $x \notin Y$, then, since $Y$ is a close subspace, we have that $dist(x, Y) = h > 0$. Then, apply Theorem 12.38 (in Bruckner's Real Analysis), and we have a functional $x^* \in X^*$ such that
$x^*(x) =h$ and $x^*(v)=0$, for all $v \in Y$. In particular, we have  $x^*(y)=0$. So $x^*(x) \ne x^*(y)$.
Remark:
For 1 above: we have that $Y = \{\alpha y: \alpha \in \Bbb R\}$. We can define a functional $g$ on $Y$ by $g(\alpha y) = \alpha$.
It is easy to see that $g$ is a linear functional in $Y$ and, since
$$ \|g(\alpha y)\| = |\alpha| = \frac{1}{\|y\|} \|\alpha y\|$$
we have that $g$ is a bounded linear functional in $Y$. Note that $g(y)=1$.
Now, applying Theorem12.29 (which is a special case of Hahn-Banach), we have that there is $l$ a bounded linear functional on $X$, that extends $g$. In particular, $l(y)=g(y)=1 \ne 0$.
A: The points $x, y\,$ are distinct, hence the span $\langle y-x\rangle =:Z\subset X\,$ is a $1$-dimensional subspace.
Define $f:Z\to\mathbb{R}\,$ to be a linear functional satisfying $f(y-x)= \|y-x\|\neq 0\,$. This completely determines $f$,
moreover $f(z)\leq\|z\|\,$ for all $\,z\in Z.$
All prerequisites are created now to apply the Hahn-Banach theorem [Thm 12.28 in Bruckner $^2\,\cdot$ Thomson] which yields an extension
$F:X\to\mathbb{R}\,$ of $f$ such that
$$F(x)\:\leq\:\|x\|\quad\forall x\in X\,.$$
This also shows the continuity of $F$. For the given points $x,y\,$ we have
$$F(y)-F(x) = F(y-x) = f(y-x) \neq 0\,,$$
thus $F$ separates $x$ and $y$.

You may also go straight to Theorem 12.37, also in
[Bruckner$\,^2\cdot\,$Thomson], and apply it with $x_0=y-x\,$.
