Existence of Bounded Linear Functional That Maps Elements of Normed Space to Distance from Subspace I'm attempting to prove the following:

Suppose X is a normed space and Y a subspace of X. Let $x ∈ X$ and denote $d$ the distance from $x$ to $Y$; that is, $d = d(x, Y ) = \inf_{y∈Y} || x − y ||$. Prove that there exists a bounded linear functional $f : X → R$ such that
  $f(x) = d$, $f(y) = 0$, for all $y ∈ Y$ and $||f||_{op} = \sup_{x \in X, ||x||=1}||fx|| ≤ 1$.

It seems like this can be proven using the Hahn-Banach Theorem, but I'm not sure how to proceed. 
Any help is much appreciated!
 A: I suppose that $d>0$.  Otherwise, the zero functional trivially solves the problem.
We apply Hahn-Banach Theorem on the subspace $\mathrm{span}\{Y,x\}=\{tx+y\in X\mid t\in \Bbb{R},y\in Y\}$ and the functional $g:\mathrm{Y,span}\{x\}\to\Bbb{R},g(tx+y)=t||x||=||tx||$ linear with respect to $t$ and $y$.  Check that $g(x) = ||x||$.  For any unit vector $u\in\mathrm{span}\{Y,x\},$ express $u$ as $tx+y$
$$|g(u)| = \left|g\left(\frac{tx+y}{||tx+y||}\right)\right|=\frac{|t|\,||x||}{||tx+y||}  = \frac{||x||}{||x-(-y/|t|)||}\le \frac{||x||}{d},$$ so $||g||_{(\mathrm{span}\{Y,x\})'} \le ||x||/d$.  (The last inequality makes use of the defintion of $d$ as $\inf\limits_{y \in Y} ||x-y||$, $Y$ as a subspace of $X$, and the assumption $d>0$.)
By Hahn-Banach, there exists a bounded functional $f(\in X')$ extending $g$ to the whole space $X$: $$||f||_{X'}= ||g||_{(\mathrm{span}\{Y,x\})'} \le \frac{||x||}{d} \text{ and } f(z) = g(z) \quad\forall\, z\in\mathrm{span}\{Y,x\}.$$
In particular, $f(x)=||x||$.  We scale this function $f$ by a factor of $d/||x||$ to finish the proof.
