Does the center of a convex region lie within that region? There's probably a simple result that says this is true, but I sure can't find it.  It seems obvious, though.
Let $D$ be a closed, compact region in $\Re^n$.  Further, let $D \subseteq [0,l]^n$ and let $B_d^\epsilon(a)$ be an $\epsilon$-ball with distance metric $d$ (e.g., the 1-norm) centered at $a \in [0,l]^n$.  Let $C$ be the non-empty intersection of $B_d^\epsilon(a)$ and $D$.  The diameter of $C$ is defined as $diam(C) = \max_{x,x' \in C} d(x,x')$. Defined analogously, $diam(B_d^\epsilon(a)) = 2 \epsilon$ and $ diam(B_d^\epsilon(a))\geq diam(C)$. Does the set of points $X^* = argmin_{x\in B_d^\epsilon(a))}[\max_{x' \in C}d(x,x')]$ include an element of $C$?  Similarly, does there exist $x \in C$ s.t. $C \subseteq B_d^{\epsilon}(x)$? Can we find a proof that this is true for C convex?
Remarks:
What we're really trying to prove here is that $C$ lies within an $\epsilon$-ball of some point in its own interior if $C$ is convex.
1) We can construct examples with $C$ non-convex, such that the intersection of $X^*$ and $C$ is empty.
2) The minimizer of $d(x,x')$ for any $x'\in C$ is $x = x'$; the objective we're trying to minimize (itself convex) is a maximum over the loss incurred by any of infinitely many convex problems, each of which has its unique minimizer in $C$.  Is there a result which says that the minimum of such a problem is attained by $x \in$ the convex hull of the solutions of the subproblems (i.e., $C$ if it is convex)? 
3) The minimizing set $X^*$ is the infinite intersection of $B_d^{r}(x) \forall x \in C$, where $r$ is the minimum value such that this infinite intersection is non-empty.  Since the individual balls are convex and compact, so is $X^*$.  Further, since $x \in C$, the intersection of each $B_d^{r}$ with $C$ is non-empty.  If we could establish (somehow via convexity of $C$) that any collection of $n$ such $r$-balls with their centers in $C$ have a non-empty intersection with $C$, we could use Helly's Theorem to prove that $C$ intersects the infinite intersection as well.
So, anyone with
a) an authoritative source which confirms one of my missing links (or just gives the main result) or
b) an independent proof?
 A: No, if I understand your question correctly.
Example: Take $d$ to be the 1-norm, as you suggest.  Let $C$ be the convex hull of the standard basis vectors.  Assume $n\ge 3$.  Then $C$ is contained in a unit ball centred at the origin, not contained in any smaller ball, and not contained in a unit ball centred at any other point — and in particular, not contained in a unit ball centred at any point of $C$ itself.
Proof: Since $C$ is $(n-1)$-dimensional, covering it by a ball (say, of radius $r\le 1$) is the same as covering it by some cross-section of a ball parallel to the affine hull of $C$.  The unit ball of the $1$-norm is the convex hull of $C$ and $-C$, which live in parallel hyperplanes, so that's the same as covering $C$ by some translate of $(1-\lambda)rC + \lambda r(-C)$ for some $\lambda\in[0,1]$.  Suppose we've done this, and suppose if possible that $\lambda>0$.  Then $C\subseteq (1-\lambda)rC + \lambda r(-C)$ (up to translation), which since Minkowski addition is cancellative implies $(1-(1-\lambda)r)C\subseteq \lambda r(-C)$, and so
$$ \frac{\lambda r}{1-(1-\lambda)r} \ge n-1 $$
(because $C$ is an $(n-1)$-dimensional simplex).  The LHS is an increasing function of $r$, its maximum value (for $r\le 1$) is $1$, so this is impossible with $n\ge 3$.  So $\lambda=0$, meaning we have to cover $C$ with the copy of $C$ that's the face of the unit ball in the positive orthant, that is, we have to centre the unit ball at the origin.
(There's probably a less cumbersome way to do this, but that's what I have off the top of my head.)
On the other hand, it's true if $d$ is the usual Euclidean norm, as arbautjc pointed out in comments.
It's also true if $C$ is convex and centrally symmetric.  (Take the origin to be $C$'s centre of symmetry; if $C\subseteq x+rB$ then $2x+C = 2x-C\subseteq x+rB$, and so $x+C = \frac12 C + \frac12(2x+C)\subseteq x+rB$, so $C\subseteq rB$.  That is, if $C$ is contained in a ball of radius $r$ then it's contained in a ball of radius $r$ centred at $C$'s centre of symmetry.)
