I believe that, at least in the case of a weakly compact convex set $K$ in a uniformly convex Banach space $X$, the *circumcenter* could be a possible good candidate. By circumcenter I mean a point which is a solution to the following minimization problem:
$$
\bar r := \inf_{x\in K} \min\{ r\geq0 \mid K\subseteq\bar B(x,r) \},
$$
where $\bar B(x,r)$ denotes the closed ball.

It is immediate to verify that the problem has a solution. Indeed, if $(x_n)_{n\in\mathbb N}\subset K$ is a minimizing sequence, that is, there exist $r_n\to\bar r$ such that $K\subseteq\bar B(x_n,r_n)$, possibly extracting a subsequence we can assume that $x_n\rightharpoonup\bar x$, then we have
$$
|y-\bar x| \leq \liminf_{n\to\infty} |y-x_n| \leq
\liminf_{n\to\infty} r_n = \bar r
\qquad \text{for all $y\in K$},
$$
which implies that $K\subseteq\bar B(\bar x,\bar r)$.

The remaining problem is the uniqueness. Assume that $x_1$ and $x_2$ are two circumcenters and set $\varepsilon=|x_1-x_2|$. By the uniform convexity, there exists $\delta>0$ such that $|u|\leq\bar r$, $|v|\leq\bar r$ and $|u-v|\geq\varepsilon$ imply $\left|\frac{u+v}2\right|\leq\bar r-\delta$. Now, if we take $y\in K$, we have $|x_1-y|\leq\bar r$, $|x_2-y|\leq\bar r$ and $|(x_1-y)-(x_2-y)|=\varepsilon$, therefore $\left|\frac{x_1+x_2}2-y\right|\leq\bar r-\delta$. But this contradicts the minimality of $\bar r$, because we have found $K\subseteq\bar B\left(\frac{x_1+x_2}2,\bar r-\delta\right)$.

To tackle the non uniformly convex case, I just have an idea, but I'm not sure if it's going to work. Maybe someone else can think about it. Given a weakly compact convex set $K_0$, we define $r_0$ as the optimal radius above and we consider the set $K_1=\{ x\in K \mid K\subseteq\bar B(x,r_0) \}$. The set $K_1$ is non-empty, convex (thanks to the convexity of the norm) and closed (I believe, by the same reasoning used for the existence). Thus, we have found another weakly compact convex set $K_1\subseteq K_0$. By iterating this argument, we find a nested family of compact sets $K_{n+1}\subseteq K_n$. Then the intersection $K_\infty=\bigcap_{n\in\mathbb N}K_n$ is non-empty. The *hope* is that one could get that $K_\infty$ is a singleton by proving something like $r_{n+1}\leq r_n/2$, or $\mathop{\mathrm{diam}}(K_{n+1})\leq\mathop{\mathrm{diam}}(K_n)/2$.

### Note to future self for improvement

We have $\mathop{\mathrm{diam}}(K_{n+1}) \leq r_n \leq \mathop{\mathrm{diam}}(K_n) \leq 2r_n$.

- The first inequality is sharp.
Example: $K_0=[0,1]\times[-1,1]\subset\mathbb R^2$ with the
$\lvert\,\cdot\,\rvert_\infty$ norm. $r_0=1$ and $K_1=[0,1]\times\{0\}$.
- The third inequality is simply useless.
- The second inequality is the one to be improved.

### Existing literature

The improvement of the second inequality is related to the Jung's constant of a metric space $(X,d)$, given by
$$
J(X) = \sup\left\{\frac{2\mathrm{rad}(A)}{\mathrm{diam}(A)} \bigg\vert A\subseteq X\right\}
$$
where
$$
\mathrm{rad}(A) = \inf_{x\in X}\sup_{y\in A} d(x,y), \qquad\qquad
\mathrm{diam}(A) =\sup_{x,y\in A} d(x,y).
$$
We have already noted that $1\leq J(X)\leq 2$. If $J(X)<2$, then we get
$$
\mathrm{diam}(K_{n+1})\leq r_n\leq J(X)/2 \mathrm{diam}(K_n),
$$
therefore
$\mathrm{diam}(K_n) \leq \left(\frac{J(X)}2\right)^n \mathrm{diam}(K_0)$
converges geometrically to $0$.

This gives us another criterion for the uniqueness of the center: it is sufficient that $J(X)<2$. For example, $J(L^\infty)=1$ [Pichugov 1988] and we have the uniqueness of the center, even though $L^\infty$ is not uniformly convex.

See also this paper and this.

### Failure in $L^1$

It is known that $J(L^1)=2$. I have found a nice example that proves this and that proves also that, unfortunately, the iterative procedure suggested above fails in $L^1$.

Consider in $L^1([0,1],\mathrm{leb}^1)$ the family $W=(w_n)_{n\in\mathbb N}$ of the wavelet functions
$$
w_n(x) = \begin{cases}
1 & \frac i{2^n}\leq x<\frac{i+1}{2^n},\ \text{$i$ even}, \\
-1 & \frac i{2^n}\leq x<\frac{i+1}{2^n},\ \text{$i$ odd},
\end{cases}
$$

It is immediate to compute $\lVert w_m-w_n\rVert_1=1$ for every $m\neq n$.
Therefore $\mathrm{diam}(W)=1$. We want to show that also $\mathrm{rad}(W)=1$. Given a function $f\in L^1([0,1],\mathrm{leb}^1)$ and $\varepsilon>0$ we can find $n\in\mathbb N$ and a function
$$
f_n = \sum_{i=0}^{2^n-1} a_i \chi_{\left[\frac i{2^n},\frac{i+1}{2^n}\right)}
$$
such that $\lVert f-f_n\rVert_1\leq\varepsilon$. (This can be done in many different ways: by hand, or using the fact that $W$ is complete in $L^2([0,1],\mathrm{leb}^1)$).
Then
$$
\lVert f_n-w_{n+1}\rVert_1 =
\sum_{i=0}^{2^n-1}
\int_{\frac i{2^n}}^{\frac{i+1}{2^n}} \frac{|a_i-1|+|a_i+1|}2 dx \geq
\sum_{i=0}^{2^n-1}
\int_{\frac i{2^n}}^{\frac{i+1}{2^n}} 1\,dx = 1,
$$
hence $\lVert f-w_{n+1}\rVert_1 \geq \lVert f-f_n\rVert_1 - \lVert f_n-w_{n+1}\rVert_1 > 1-\varepsilon$.
This proves that a ball containing $W$ must have radius at least $1$.

Furthermore, if we take $K=\mathrm{co}(W)$, then the ball of radius $1$ covering $K$ can be centered at any point of $K$, which means that $H=K$ and the procedure described above doesn't converge to a single point.