I bought a book called "Problems in Real and Functional Analysis" because it contained solutions. There is one problem I couldn't quite convince myself of, but when I looked for the solution it was missing:
Problem 8.30. Let $Y$ be a proper closed subspace of a normed linear space $X$. Prove
$$ \sup_{0 \neq x \in X} \frac{d(x,Y)}{||x||} = 1 $$
Attempt: Case 1: If $x \in Y$ then $d(x,Y)=0$ and $\frac{d(x,Y)}{||x||} = 0 \leq 1$.
Case 2: If $x \in X\backslash Y$ then $d(x,Y)>0$ because $Y$ is closed. Thus for some $y \in Y$ we have $ d(x,Y)= ||x-y|| $.
Suppose $x-y \in Y$ then $\exists y_0 \in Y$ such that $x-y=y_0$ but then $x=y_0 + y$ which contradicts the closure of $Y$. Therefore $x-y \in X\backslash Y$ and so $\exists x_0 \in X\backslash Y$ such that $x-y = x_0$ or in other words $x=x_0 +y$ and $-y = x_0 -x $. Clearly $x_0$ is nonzero because $0 \in Y$. Then
$\frac{d(x,Y)}{||x||} = \frac{ ||x-y||}{||x||} = \frac{||x +(x_0 - x)||}{||x_0 +y||} = \frac{||x_0||}{||x_0 + y||} \leq 0$
For this fixed $y$ get a sequence $x_n$ in $X\backslash Y$ such that $d(x_n,Y) = ||x_n-y||$ as well as $||x_n-y||< ||x_{n+1} - y ||$ (can I assume this?). For each $x_n-y$ we can say $x_n - y = x_0^n$ for some $x_0^n \in X$. Notice $||x_0^n|| < ||x_0^{n+1}||$ as well. Then
$$\lim_{n \rightarrow \infty} \frac{||x_0^n||}{||x_0^n + y||}=1 $$
and hence
$$ \sup_{0 \neq x \in X} \frac{d(x,Y)}{||x||} = 1 $$
Note that $d(x + y, Y) = d(x, Y)$ for $x \in X$ and $y \in Y$. For $x \in X \setminus Y$, and $\epsilon > 0$, there exists $y \in Y$ such that $$ ||x + y|| (1 - \epsilon) < d(x, Y), $$ so $$ 1 - \epsilon < \frac{d(x, Y)}{||x + y||} = \frac{d(x', Y)}{||x'||}, $$ with $x' = x + y$. Thus, more exactly than the original statement, for any given $x \in X \setminus Y$ $$ 1 = \sup\{ \frac{d(x', Y)}{||x'||} : x' \in x + Y\}. $$