 Why the 2-core is unique

Suppose $T$ is any vertex set whose induced subgraph has every degree of a vertex at least two. No vertex of $T$ can be the first one deleted from $T$: before such a deletion it still has its two graph neighbours within $T$, contradicting eligibility for deletion. Consequently every qualifying $T$ survives every legal deletion sequence. Once deletion stops, its remainder $R$ itself qualifies, because every surviving vertex has at least two surviving graph neighbours. Therefore $T\subseteq R$ for every qualifying $T$: $G[R]$ is the unique maximum and the final remainder is independent of deletion order. The intermediate FIFO queues and deletion orders can differ.

There is also a direct existence argument. If $A$ and $B$ qualify, then $A\cup B$ qualifies: each vertex retains at least the two graph neighbours it already had in $A$ or in $B$. Taking the union of all qualifying vertex sets produces the maximum. This rules out a collection of competing, incomparable inclusion-maximal answers. It also explains why restricting the search to connected subgraphs would change the question.