Let $A$ be normed vector space, and $B\subsetneq A$ a linear subspace that is not necessarily closed. Let $\epsilon>0$.
There exists a $a\in A$ s.t. $||a||=1$ and $d(a,B)=\inf_{b\in B}||a-b||>1-\epsilon$.
Is this statement true or false? My intuition says false, but I would have no idea how to find a counter-example (or proof if it's true). Can somebody give me a hint how to start (and tell me whether it is true or false)?
For $\epsilon > 1$, every $a\in A$ with $\lVert a\rVert = 1$ has the desired property, since clearly
$$d(a,B) = \inf_{b\in B} \lVert a-b\rVert \geqslant 0 > 1 - \epsilon.$$
For $0 < \epsilon \leqslant 1$, such an $a$ exists if and only if $B$ is not dense in $A$, i.e. $\overline{B} \neq A$.
That follows from the fact that $d(a,M) = d(a,\overline{M})$ for all subsets $\varnothing \neq M \subset A$ and $a \in A$, and Riesz' lemma. (The restriction $M \neq \varnothing$ is not actually necessary if one uses the [useful and natural] convention $d(a,\varnothing) = +\infty$.)
The inequality $d(a,N) \leqslant d(a,M)$ for $M \subset N$ immediately yields $d(a,\overline{M}) \leqslant d(a,M)$. For the inequality in the other direction, pick an arbitrary $x \in \overline{M}$. Then there is a sequence $(x_k)_{k\in \mathbb{N}}$ with $x_k \in M$ for all $k$ and $x_k \to x$. By the continuity of the distance function, it follows that
$$d(a,x) = \lim_{k\to \infty} d(a,x_k) \geqslant \inf \{d(a,m) : m \in M\} = d(a,M).$$
This holds for all $x \in \overline{M}$, whence
$$d(a,\overline{M}) = \inf \{ d(a,x) : x \in \overline{M}\} \geqslant d(a,M),$$
and the equality $d(a,\overline{M}) = d(a,M)$ is established.
Now if $\overline{B} = A$, then we clearly have
$$0 \leqslant d(a,B) = d(a,\overline{B}) = d(a,A) = \inf \{ \lVert a-x\rVert : x \in A\} \leqslant \lVert a - a\rVert = 0,$$
and consequently $d(a,B) = 0 \leqslant 1-\epsilon$ for all $a \in A$, so in that case no $a\in A$ with the desired properties exists.
And if $\overline{B} \neq A$, then $\overline{B}$ is a proper closed subspace of $A$. Choose an arbitrary $a_0 \in A \setminus \overline{B}$. Since $\overline{B}$ is closed, there is an $r > 0$ such that the open ball with radius $r$ and centre $a_0$ doesn't intersect $\overline{B}$, whence $d(a_0,\overline{B}) \geqslant r > 0$. Let $a_1 = \frac{1}{d(a_0,\overline{B})}\cdot a_0$. Then $d(a_1,\overline{B}) = 1$, and by definition of the infimum, there is a $b_1 \in \overline{B}$ with $\lVert a_1 - b_1\rVert < \frac{1}{1-\epsilon}$. Then
$$a := \frac{1}{\lVert a_1 - b_1\rVert}\cdot (a_1 - b_1)$$
has the desired properties. $\lVert a\rVert = 1$ is clear, and for every $b \in \overline{B}$ we have
$$\lVert a - b\rVert = \frac{1}{\lVert a_1 - b_1\rVert}\bigl\lVert a_1 - \underbrace{(b_1 + \lVert a_1 - b_1\rVert\cdot b)}_{= b' \in \overline{B}}\bigr\rVert \geqslant \frac{1}{\lVert a_1 - b_1\rVert} d(a_1,\overline{B}) = \frac{1}{\lVert a_1 - b_1\rVert},$$
so
$$d(a,B) = d(a,\overline{B}) \geqslant \frac{1}{\lVert a_1 - b_1\rVert} > 1-\epsilon.$$