Given a set $S$ of non-zero vectors in $\mathbb{R}^n,$ and a subspace $L,$ consider $f(S,L)=\max_{s \in S}\frac{\|Ps\|}{\|s\|}$$f(S,L)=\max_{s \in S}\frac{\|Ps\|_2}{\|s\|_2}$ where $Ps$ is the orthogonal projection of $s$ onto $L.$
If that seems too broad, then the case $d=1$ would be of interest.
Summary
There are several good answers here. It is a toss-up which one to choose. I'm going to repost the $d>1$ case as a new question.
Here is a somewhat selective review, all for the case $d=1.$
Gerhard mentioned that $(1,-1,0,0,\cdots)$ attains $\sqrt{1/2}.$
Seva suggested that $(1,1/\sqrt{2},1/\sqrt{3},\cdots)$ is asymptotically optimal attaining $O(\sqrt{1/\log n}).$ This is indeed better but the smallest $n$ for which it beats $\sqrt{1/2}$ is (by my calculations) $n=1203.$ The nice optimality proof references $(1,\sqrt{2}-1,\sqrt{3}-\sqrt{2},\cdots)$ This is actually a better choice, it first beats $\sqrt{1/2}$ for $n=56.$ The two choices are comparable (up to scalar multiple) because $\sqrt{k+1}-\sqrt{k} \approx 1/2\sqrt{k}$. The approximation is pretty good, except for $k=0.$ As this is the largest entry, it takes longer to beat $\sqrt{1/2}$.
Denis essentially does mention $(1,\sqrt{2}-1,\sqrt{3}-\sqrt{2},\cdots)$ and, at least in one edit, suggests reflecting in the center: So for $n=6$ use $(a_1,a_2,a_3,-a_3,-a_2,-a_1)$ where $a_k=\sqrt{k}-\sqrt{k-1}.$ This is superior at $n=4$ and even at $n=3$ with the right rule for the center: It seems obvious that there should be symmetry and entries summing to $0$, but actually $(1,\sqrt{2}-1,-1)$ is the best choice for $n=3$.
Emil nicely lays this out in a comment. Here is my somewhat informal explanation (with some minor details skipped) , in case another perspective is useful. I suggest at the end that this should remind one of linear programming.:
There is no harm in assuming that the subspace is generated by a vector $v=(v_1,v_2,\cdots,v_n)$ with $v_1 > v_2 > \cdots > v_n$ (actually, only $\ge$ is clear, but I will ignore that in the interests of brevity.). By a mild abuse of notation, for $j>0$ let $x_{j}$ be the vector whose first $j$ entries are $1$ and the rest $0$ while $v_{-j}$ has first $n-j$ entries $0$ and the last $j$ equal to $1$.
Given $v$, Let $B \subset X$ be the set of vectors in $X$ which attain the maximum of $\frac{\|Px\|}{\|x\|}.$ I make two claims about $B$:
- Every vector in $B$ is $x_j$ for some $j$ with $v_j > 0$ or else $x_{-j}$ for some $j$ with $v_{n-j} < 0$ (think about why any other vector with $j$ entries equal to $1$ will be worse.)
- Unless $B$ is a basis of $\mathbb{R}^n$, There is a $v$ which gives a better ratio. (Because otherwise we should be able perturb the entries of $v$ in such a way as to lower $\frac{\|Px\|}{\|x\|}$ for the vectors already in $B$ (simultaneously keeping them equal and maximal) while raising that ratio slightly for some $x_j$ not yet in $B$.
Together these tell us that $B=\lbrace x_{1},x_{2},\cdots x_n\rbrace$ or else $B=\lbrace x_{1},x_{2},\cdots x_q;x_{q-n},\cdots,x_{-2},x_{-1}\rbrace$ where $v_q>0>v_{q+1}.$ Since the various ratios must all be the same, we deduce that $v=(\alpha a_1,\alpha a_2, \cdots,\alpha a_q; -\beta a_{n-q},\cdots,-\beta a_2,-\beta a_1)$ for the $a_k$ as above. Here $\alpha$ and $\beta$ are positive constants. A bit of reflection shows that they must be equal and hence can be taken to be $1$. Finally, $q$ should be $ n/2 $ (rounded if needed).
Comment: (disclaimer: I am an optimist but not an optimizer so the following may inexact.) Here we have the convex optimization problem of choosing $v$ so as to minimize the largest of $\frac{\|Px\|_2}{\|x\|_2}.$ Had it been $\frac{\|Px\|_1}{\|x\|_1}$ we would have been able to use linear programming. My argument above has the feel of linear programing, perhaps using duality. Perhaps a similar method could uncover optimal subspaces of dimension $d \ge 2.$
