How not to compute the smallest circle enclosing a finite set of circles
This step of removing $X$ from $L$ before continuing recursion actually improves the algorithm, because it removes the already-added $X$ from the pool of basis candidates. It will always, provably work, because it's equivalent to the existing algorithm, and it will also work for higher dimensions.

Trace through the algorithm. When you use $MSW(L,B^{\prime\prime})$, there is $X \in L$ and $X \in B^{\prime\prime}$. Suppose we chose it again in step 2. Regardless of the result of step 3, $B^\prime$ will always have $X$, because the recursive function has the invariant $B \subseteq MSW(L,B)$.

In other words, your improvement to the algorithm shortcuts to step 3 in the part where $X$ is chosen.