Convexity of open balls in a metrizable subset of a locally convex t.v.s. The following result is well known (Rudin, Functional Analysis, Theorem 1.24): If $X$ is a locally convex topological vector space with a countable local base then there is a compatible metric $d$ such that all open balls are convex.
My question is: Assume that $X$ is a locally convex t.v.s. which is not metrizable and $C$ is a compact, convex and metrizable subset of $X$. Is there a metric compatible with the topology of $C$ such that all open balls are convex?
 A: Let
$
  s: X\times X\to X
  $
be the subtraction map, namely
$$
  s(x,y) = x-y, \quad \forall x,y\in  X,
  $$
and put
$$
  D:= s(C\times C) =  \{x-y: x, y\in  C\}.
  $$
We next claim that $D$ is homeomorphic to a quotient of $C\times C$.  Indeed, consider the equivalence relation $R$ on $C\times C$
according to which two points are equivalent iff they have the same image under $s$.   Since $s$ obviously respects $R$,  there is a unique
continuous map
$$
  \tilde s:(C\times C)/{R}\ \to \ D,
  $$
such that $s=\tilde s\circ  q$, where $q$ is que quotient map.
It is not hard to see that $\tilde s$ is a bijective map,
and since $(C\times C)/{R}$ is compact and $D$ is Hausdorff, we conclude that $\tilde s$ is a homeomorphism, thus proving the claim.
Every Hausdorff quotient of a compact metrizable space is itself metrizable, so $D$ is metrizable and hence also   first countable.
Noticing that  $0$ lies in $D$,  we may then find a countable open neighborhood base
for $0$ in the relative topology of $D$,  say  $\{V_n\}_n$.   Write each $V_n$ as $W_n\cap D$,  where $W_n$ is an open subset of
$X$, and choose an open,  convex,  balanced set $U_n$,  with
$$
  0\in U_n\subseteq W_n.
  $$
It is then evident that
$$
  V'_n := U_n\cap D \subseteq  W_n\cap D = V_n,
  $$
so the $V'_n$
form an open neighborhood base for $0$ in the relative
topology of $D$.
For each $n$, let $p_n$ be the Minkowsky functional associated to $U_n$ and consider the (not necessarily
Hausdorff), locally convex topology $\tau'$ on $X$ generated by the $p_n$.
We next claim that $(C, \tau ')$ is Hausdorff.  To see this, let $x$ and $y$ be  distinct points  in $C$,
and notice that
$$
  z:= y-x\in  C-C = D.
  $$
Choose an open set $A\subseteq X$, such that $0\in A$ and
$z\notin A$,  and  let $n$ be such that $V'_n\subseteq  A\cap D$.
Notice that $z\notin U_n$ since otherwise
$$
  z\in  U_n\cap D = V'_n\subseteq  A\cap D\subseteq A.
  $$
It follows that
$$
  1\leq p_n(z)=p_n(y-x),
  $$
so $x$ and $y$ may be separated in the topology generated by $p_n$, and hence also in $\tau '$, proving our
claim.
Clearly $\tau '$ is coarser than the default
topology $\tau $ of $X$,   so the identity  map
$$
  \text{id} : (C, \tau ) \to  (C, \tau ')
  $$
is continuous.
Given that $(C, \tau)$ is compact, and that we now know that $(C, \tau ')$ is Hausdorff, we have that
the above map is a homeomorphism.
In case $(X,\tau ')$ happens to be Hausdorff, an application of the result by Rudin mentioned by the OP would provide the
desired metric, but a slight adaptation of it also works even if $\tau '$ is not Hausdorff.  By that I mean that Rudin's
proof carries through without the assumption that $(X,\tau ')$ is Hausdorff, as long as we drop the
requirement that $d$ satisfies
$$
  d(x,y)=0\Rightarrow x=y,
  $$
that is, as long as we satisfy ourselves with a pseudo-metric, rather than a metric.
The resulting pseudometric $d$ would of course be compatible with the topological space  $(C, \tau ')$, and hence also $(C, \tau )$,
but since this is a Hausdorff space,  as seen above,  $d$ must be a metric when restricted to $C$.
A: $Y=\operatorname{span}(C)$ is a locally convex tvs which is still metrisable I think (using that $C$ has a countable base, being compact metrisable). So the first result would apply there.
