It is on my mind that a related region in $X$ constructed from $K$ is always compact, I think. Any measure $\mu$ on $K$ has a center of gravity $c$, defined as the minimum $x$ of the average value of $d(x,y)^2$ with $y$ sampled from $\mu$. The set of Borel measures on $K$ is compact by the Banach-Alaoglu theorem in the weak-* topology, and my intuition is that the position of the center of gravity is continuous with respect to this topology. (If this intuition is wrong, then the rest of this post is not all that interesting.) This is not the same as the convex hull, but it seems interesting as a possible approximation.

Any point in the strict convex hull of $K$ (not its closure) is reached by a sequence of binary operations on pairs of points from a finite list. The initial list of points is in $K$, and then certain pairs $x$ and $y$ to make new points $z$. By definition, $z$ is at distance $p$ along the way from $x$ to $y$. This description induces a measure on the initial set of points. For instance suppose that we start with $x_1, x_2, x_3, x_4 \in K$, then take the point $y_1$ that is $p_1$ along the geodesic from $x_1$ to $x_2$ and the point $y_2$ that is $p_2$ along the geodesic from $x_3$ to $x_4$. Then finally the point $z$ is $q$ along the way from $y_1$ to $y_2$. The induced measure on the original list of points is then $qp_1[x_1]+q(1-p_1)[x_2]+(1-q)p_2[x_3]+(1-q)(1-p_2)[x_4]$. This measure has a center of gravity $c$, and I am wondering how far away $c$ can be from $z$. If $X$ happens to be a vector space, then $c=z$, but in general they are not equal.

There is a mutual generalization of points in the convex hull and centers of gravity. Starting with a base list of points $x_1,\ldots,x_n \in K$, there is a $k$-ary operation with weights that replaces $k$ of the points with their center of gravity. If these operations are repeated in the pattern of a weighted tree $T$, then the computation produces a point $z$ which could be in the convex hull (if $T$ is binary), or could be a center of gravity (if $T$ is a shrub), or could be various things in between. Now suppose that $T$ is a complicated tree. We can flatten it to make it a shrub $T_1$ that yields a point $z_1$. Then in various ways we can unflatten $T_1$, step by step, to approach $T = T_n$. Assuming the first paragraph, the points that can be reached by trees of bounded depth are a compact set. I do not know enough about $\text{CAT}(0)$ spaces to draw any conclusions, but it seems possible that the points $z_k$ approach the final point $z = z_n$ quickly enough to establish compactness. Or if this does not happen, then that could be evidence against compactness of the convex hull.

(To be clear, this is just a proposal and not a solution.)

Here another way to state the proposal without any direct use of center of mass, although it is still suggested by the fact that the set of centers of mass is compact.

For any $0 \le p \le 1$, there is a binary operation $x \heartsuit_p y$ on points in $X$. By defintion, $x \heartsuit_p y$ is the point $z$ such that $d(x,z) = pd(x,y)$ and $d(y,z) = (1-p)d(x,y)$. Let $x_1,\ldots,x_n$ be a list of points in $K$, possibly with repetitions. Then every word $w$ in the points of $K$ written in this notation defines a point $z \in X$. We can also compute the same word in the vertices of a Euclidean simplex $\Delta_{n-1}$. We thus obtain a continuous map $f_T:\Delta_{n-1} \to X$ that only depends on the tree structure $T$ of $w$. How far away are these maps from each other for two different trees? (There are $(2n-3)!!$ distinct trees.) If you have enough control over the distance, then the closed convex hull of $K$ is compact. More precisely, the hope is to find a compactification of the space of words such that the evaluation map extends continuously.

For example, if $n=3$, we can define three tree centers of three points $x,y,z$, namely $x \heartsuit_{1/3} (y \heartsuit_{1/2} z)$ and its cyclic permutations. How far apart can they be in a $\text{CAT}(0)$ space? For instance, in a tree, which is one kind of opposite to a Euclidean space, the tree centers of a unit equilateral triangle are at most 1/3 away from each other.

In the comments to my other post on possible counterexamples, Anton also asks for references. I found the paper Nonexpansive retracts in Banach spaces, by Kopecká and Reich. They say,

This proof of Theorem 2.10 works equally well in any Hadamard space in which the closed convex hull of a finite number of points is compact. It follows then that the Plateau problem can be solved in such spaces. Unfortunately, it is not known which Hadamard spaces have this property. However, it is shown in [We, Theorem 1.6] that Plateau’s problem can be solved in every Hadamard space (regardless of whether it has this property or not)."

Their theorem 2.10 is an interesting, stronger property that they show follows from Anton's property: Every compact set $K$ is contained in a compact, 1-Lipschitz retract of the space $X$. Clearly such a retract is also convex, so the convex hull of $K$ inside is compact. Moreover, the subject of their Theorem 2.10 is exactly my example 2, so that example does not work. Moreover, Anton's property already has a name in the literature, what they call CNEP, and they reduce to the case that $K$ is finite. Finally, as of 2006, these authors describe it as an open problem.