Solution:

Here is a sequence of 3 moves that works: $54321 \rightarrow 32541 \rightarrow 34125 \rightarrow 12345$.

But how do we know we can't do it in 2 moves? From any position there are 20 possible permutations via block moves, 16 from moving a block of size 1 and 4 from moving a block of size 2. One could simply write the 20 permutations of $54321$ and the 20 permutations of $12345$ and try to see that they have nothing in common, which would suffice since the inverse of a block move is also a block move. A more clever method is to notice that if we could sort $54321$ in 2 moves then we could sort $4321$ in 2 moves as well by simply deleting the 5 from each step. But $4321$ has only 10 permutations from block moves, namely $3421,3241,3214,4231,4213,2431,4312,1432,4132$, and $2143$. The 10 permutations of $1234$ are $2134,2314,2341,1324,1342,3124,1243,4123,1423$, and $3412$. These two sets of permutations have nothing in common, thus it takes at least 3 moves to sort $4321$, and hence at least 3 moves to sort $54321$.

Solution:

There is a more elegant way to show we need at least 3 moves. Given a permutation of $\{1,2,3,4,5\}$ (or any ordered set), define a descent to be an adjacent pair of numbers in the permutation such that the left number is greater than the right one. For example, $12345$, $34215$, and $54321$ have $0$, $2$, and $4$ descents, respectively. Any permutation obtained from $12345$ by one block move has (at most) one descent, at the left edge of the moved block. Similarly, any permutation obtained from $54321$ by one block move has (at least) three descents, so that we can't get from $54321$ to $12345$ by two block moves.