Given any two distinct elements $x,y \in X$, prove that there exists a bounded linear map $f: X \to \mathbf{C}$ with $\|f\| = 1$ and $f(x) \ne f(y)$.
My attempt: by a corollary of Hahn-Banach, we have that for each $x \in X$ there is some $\ell_{x} \in X^{*}$ such that $\|\ell_{x}\|=1$ and $\ell_{x}(x)=\|x\|$. So we already have a bounded linear map satisfying the first property but how do I go about showing that $\ell_x(x)\ne \ell_x(y)$?
Let $M=\Bbb C(x-y)$ and let $\varphi\colon M\longrightarrow\Bbb C$ the linear map defined by $\varphi\bigl(\lambda(x-y)\bigr)=\lambda$. Then $\varphi$ is continuous and therefore, by the Hahn-Banach theorem, you can extend it to a $F\in X^*$. Then $F(x-y)=1$. So, take $\frac F{\|F\|}$.