On the proof that Let $X$ be a normed vector space where $x,y\in X:x\neq y$. Show that $\exists f\in X^\ast:f(x)\neq f(y)$ Problem: Let $X$ be a normed vector space where $x,y\in X:x\neq y$. Show that $\exists f\in X^\ast:f(x)\neq f(y)$.
Attempt: Take $z\in X:z=y-x$, where $x\neq y\neq 0$. Then, $\|z\|=\|y-x\|=\delta\gt0$. If $Y=\{0\}$ then $\|z\|=\|x\|=\delta$.
Then we can make use of one of the corollaries to the Hahn-Banach theorem, namely,
"If $X$ a normed vector space and if $x\neq0,\,\exists\, f\in X^*:f(x)=\|x\|$ and $\|f\|=1$"
Using this we have that,
$$f(z)=f(y-x)=\|x\|=\delta$$
$$\iff f(y)-f(x)=\delta$$
$$\iff f(y)=f(x)+\delta$$
So that then, clearly, $f(y)\ge f(x)$ which is sufficient to say that $f(y)\neq f(x)$ as required.
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I want to clarify the reasoning and justification of the above proof. 


*

*The corollary to the Hahn Banach theorem we used above is really a corollary of a corollary. It follows from setting $Y=\{0\}$ in this here:
"Let $X$ be a normed vector space, $Y\subseteq X$ a linear subspace, and suppose that $\delta=\inf_{y\in Y}\|x-y\|\gt0$. Then $\exists\,f\in X^*:f(x)=\|x\|$ and $\|f\|=1$".
Given that we take $z=y-x$, so that it follows, $\|z\|=\|y-x\|=\delta\gt0$, it seems implied that $y\in Y$, so that when we set $Y=\{0\}$ we get that $\|z\|=\|x\|=\delta$. But then, why don't we set $y=0$ in what follows afterwards, namely, in $f(z)=f(y-x)$?

*Consulting several different sources, it seems to me that $Y\subseteq X$ in the main Hahn-Banach corollary (that stated in 1. above) should be closed. Is this correct, and should it be stated? Otherwise would it not follow that $\delta=\inf_{y\in Y}\|x-y\|\geq0$?
 A: You probably mean this corollary:

Let $X$ be a normed space, $Y \subseteq X$ its subspace, and $x \in X$ such that $\delta = \inf_{y \in Y}\|x-y\| > 0$. Then there exists a bounded linear functional $f \in X^*$ such that $\|f\| = 1$, $f(x) = \delta$ and $f|_Y = 0$.

Notice that by demanding $d(x, Y) = \delta > 0$ you are in fact stating that $x \notin \overline{Y}$. So you can either assume that $Y$ is closed and take $x \notin Y$, or just take $x \notin \overline{Y}.$
Your first corollary 

Let $X$ be a normed space and $x \in X, x \ne 0$. Then there exists $f \in X^*$ such that $\|f\| = 1$ and $f(x) = \|x\|$.

is now indeed a consequence of the former one:
Take $x \in X, x \ne 0$. Consider $Y = \{0\}$ and notice $x \notin \{0\} = \overline{Y}$. We have $\delta = d(x, \{0\}) = \|x\|$. The previous corollary implies that there exists $f \in X^*$ such that $\|f\| = 1$, $f(x) = \delta = \|x\|$ and $f|_{\{0\}} = 0$, so $f$ is precisely the desired functional.
Using this, we can show the desired claim, as suggested by @Idontknow:
Take $x, y \in X$, $x \ne y$. We have $x - y \ne 0$ so there exists $f \in X^*$, such that $\|f\| = 1$ and $f(x) - f(y) = f(x - y) = \|x - y\| \ne 0$. It follows that $f(x) \ne f(y)$.
Notice that you cannot claim $f(x) > f(y)$ since those might be complex numbers.
