My question is, how can DES exhibit such behaviour if the key space is about $2^{56}$ and block space is $2^{64}$
Let us assume that the attacker has a single plaintext/ciphertext pair, that is, two eight byte values $P, C$ with $C = \text{DES}_k(P)$, where $k$ is the correct unknown key.
Then, let us consider the values $\text{DES}_{k'}(P)$; there $k'$ ranges over the $2^{56}-1$ possible incorrect keys. If we model DES with the incorrect key as a random permutation, then what we get is a list of $2^{56}-1$ random 8 byte values, each one of which has a $2^{-64}$ probability of just happening to be $C$.
Hence, the expected number of times the value $C$ appears on that list is $(2^{56}-1)2^{-64} \approx 2^{-8}$ (and the probability that the value $C$ appears at least once on the list is a tad smaller).
Hence, there does indeed exist a nontrivial probability that a brute force search would find two keys; the correct key $k$, and another key $k'$ that just happens to map $P$ to $C$.
Of course, in practice, we never really get only one plaintext block and one ciphertext block; we generally get additional ciphertext blocks that, at the very least, attempt to decrypt, and see if they make sense; that'll allow us to distinguish the correct key from any incorrect ones.