Best approximation to reflexive subspace - converse Let $X$ be a normed vector space.
I know if $X$ is reflexive, then for every normed space $Y$ that $X$ is a subspace of $Y$, $X$ has a best approximation to every $y \in Y$ (which is to say there exists $x\in X$ that $d(y, X) = \|x-y\|$).
I need to prove the converse of the above fact. To be more precise, if for every normed space $Y$ that $X \subset Y$, $X$ has a best approximation to every $y \in Y$, is it true that $X$ is reflexive?
 A: $\newcommand{\dist}{\text{dist}}
\newcommand{\R}{\mathbb R}
\newcommand{\diam}{\text{diam}}
\newcommand{\and}{\quad \text{and} \quad}
$The conjecture proposed by the OP is  true:  every non-reflexive space may be embedded into a larger space $Y$ in such
a way that the minimization problem
$$
  \|y-x_0\| = \inf\{\|y-x\| : x \in X\}
  $$
admits no solution for some $y$ in $Y$.   Proving this conjecture obviously entails constructing the larger space $Y$,
so it is necessarily a somewhat long argument.
Lemma.  Let $X$ be a normed space and let $C$ be any nonempty convex subset of $X$ whose  diameter satisfies
$$
  \diam(C)\leq 2.
  $$
Then $X\oplus \R$ becomes a normed
space if equipped with the norm
$$
  \|(x,a)\| = |a| + \dist(x,aC),\quad \forall x\in X,\quad \forall a\in \R.
  $$
In addition, the correspondence  $x\mapsto (x, 0)$ defines  an isometry from $X$ onto a closed hyperplane of $X\oplus \R$.
Proof. It is obvious that
$\|(x,0)\| = \|x\|$,
so the isometry claim is clear.
We leave it for the reader to prove that
$$
  \|\lambda (x, a)\| =   |\lambda |\|(x, a)\|, \and   \|(x, a)\| =  0 \Rightarrow (x, a)=0,
  $$
and we concentrate instead in  proving  the triangle inequality
$$
  \|(x_1+x_2,a_1+a_2)\| \leq    \|(x_1,a_1))\| +  \|(x_2,a_2)\|,
  \tag 1
  $$
which we will first verify
in the special case that $a_1$ and $a_2$ have the same sign.
If both $a_1$ and $a_2$ are zero, the claim follows trivially, so we assume that $a_1+a_2\neq 0$.
Fixing any $\varepsilon >0$, choose $y_i$ in $C$, for each $i=1,2$, such that
$$
  \|x_i-a_iy_i\|<\dist(x_i,a_iC)+\varepsilon .
  $$
Then
$$
  y:= \frac{a_1}{a_1+a_2}y_1 + \frac{a_2}{a_1+a_2}y_2
  \tag 2
  $$
lies in $C$, whence
$$
  \dist(x_1+x_2,(a_1+a_2)C) \leq
  \|x_1+x_2-(a_1+a_2)y\| = $$$$ =
  \|x_1+x_2-(a_1y_1+a_2y_2)\| \leq
  \|x_1-a_1y_1\| +   \|x_2-a_2y_2\| < $$$$ <
  \dist(x_1,a_1C) + \dist(x_2,a_2C) + 2\varepsilon .
  $$
Combining this with the fact that $|a_1+a_2| \leq |a_1| +|a_2|$, we have proven (1).
Let us now prove (1) in the remaining case that $a_1$ and $a_2$ have opposite signs.   Unlike the previous case the
above argument employing (2) does not work because $y$ will not be in $C$.    It is here that the bound on the diameter of $C$ becomes relevant.
Interchanging $(x_1,a_1)$ and $(x_2,a_2)$ if
necessary, we may assume that $|a_1|\geq |a_2|$.  Furthermore, since we already know  that $\|\cdot\|$ is homogeneous, we may
multiply $(x_1,a_1)$ and $(x_2,a_2)$  by $-1/a_2$ and hence assume that $a_2=-1$.  Finally, given that $a_1$ is
positive and $|a_1|\geq 1$, we may write $a_1=1+b$, with $b\geq 0$.  Therefore the two elements we are working with may now be
represented by
$$
  (x_1,1+b)\and (-z_2,-1),
  $$
where we made a last minute  parameter change, namely $x_2=-z_2$.
Unraveling (1)  with these assumptions we see that we need to prove that
$$
  \dist(x_1-z_2,bC)\leq  2 + \dist(x_1,(1+b)C)+ \dist(z_2,C).
  \tag 3
  $$
Fixing any $\varepsilon >0$, choose $y_1$ and $y_2$ in $C$ such that
$$
  \|x_1-(1+b)y_1\|< \dist(x_1,(1+b)C) +\varepsilon ,
  $$
and
$$
  \|z_2-y_2\|< \dist(z_2,C) +\varepsilon .
  $$
Then
$$
  \dist(x_1-z_2,bC)\leq
  \|x_1-z_2 - by_1\| = $$$$ =
  \|x_1-(1+b)y_1- (z_2 -y_2) + y_1 - y_2\|\leq  $$$$ \leq
  \|x_1-(1+b)y_1 \|+\|z_2 - y_2\| + \|y_1 - y_2\|\leq  $$$$ \leq
  \dist(x_1,(1+b)C) + \dist(z_2,C) +2\varepsilon  + \diam(C).
  $$
Since $\varepsilon $ is arbitrary and $\diam(C)\leq 2$, we see that (3) is satisfied.
QED

This said, let us be given a non-reflexive space $X$.  By Smulian's Theorem (see below) choose a nested sequence $C_1
\supseteq C_2 \supseteq C_3 \supseteq \cdot \cdot \cdot \ $ of nonempty closed bounded convex subsets with empty intersection.   By
scaling everything by the same fixed factor we may assume that each $C_n$ is contained in the unit ball of $X$, and
hence, in particular, $\diam(C_n)\leq 2$.
For each $n$ let $\|\cdot\|_n$ be the norm on $X\oplus \R$ constructed in the Lemma relative to $C_n$, and define a new norm on
$X\oplus \R$ by
$$
  \|(x, a)\| =
  \sum_{n=1}^\infty   \frac{\|(x, a)\|_n}{2^n}.
  $$
Since  $\|(x, 0)\|_n = \|x\|$, for every $n$ and every $x\in  X$, and since   $\sum_{n=1}^\infty   1/2^n=1$, we see that  $\|(x,
0)\| = \|x\|$, so
$X$
also embeds into $X\oplus \R$ relative to the above norm.
We now claim that there is no point in $X$ (or rather in its canonical image in $X\oplus \R$) minimizing the distance to the
vector $u:= (0,-1)$.  To see this notice that for every $x$ in $X$ one has that
$$
  \|x-u\|_n =
  \|(x, 0)-(0, -1)\|_n =
  \|(x, 1)\|_n =
  1 + \dist(x,C_n) \geq  1,
  $$
with equality iff $x\in C_n$.  Therefore
$$
  \|x-u\| \geq
  \sum_{n=1}^\infty  \frac 1{2^n} = 1,
  $$
so we see that
$\dist(u,X)\geq 1$.
On the other hand, if $x\in  C_k$, then
$$
  \|x-u\| =
  1 +   \sum_{n=k+1}^\infty  \frac{\dist(x,C_n) }{2^n}
  $$
which can be made as close as desired to 1 if $k$ is large enough, meaning that in fact   $\dist(u,X)=1$.
The punch line is then that there is no $x$ in $X$ such that
$$
  \|x-u\| =   1 +   \sum_{n=1}^\infty  \frac{\dist(x,C_n) }{2^n} = 1,
  $$
since such an $x$ needs to be in every $C_n$, but we know that $\bigcap_nC_n$ is empty.

Not many people read Russian, so here is the statement:
Theorem (Smulian [1]) A normed space $X$ is reflexive iff every nested
sequence $C_1 \supseteq C_2 \supseteq C_3 \supseteq \cdot \cdot \cdot \ $ of nonempty closed
bounded convex subsets of $X$ has nonempty intersection.
[11] V.L. Smulian, On the principle of inclusion in the space of the type (B), Mat. Sb. (N.S.) 5 (1939) 317-328.
A: A proof can be found here (page 399)
