Assume we have a deterministic infinite automaton. $DIA = (Q, δ, q_0, F)$. Meaning there is no limit to the number of states. I want to prove or disprove that this module equals a normal DFA.
Attempt
I tried to disprove it using the language L = {$a^nb^n|0 < n ∈ N$}, but I am not sure if this is enough to disprove it.
That approach indeed works. It is well known that $L = \{a^n b^n \mid n \in \mathbb N\}$ is not regular and hence not recognized by any "normal" DFA, so it suffices to show that $L$ can be recognized by a DIA. I'll leave the details to you, but using infinite states we can remember the number of $a$ we read so far and then count down when reading $b$.
Indeed, one can show that every language can be recognized by a DIA: Let $L$ be some language and consider the DIA $\mathcal A_L$ with state set $Q = \{q_w \mid w \in \{0, 1\}^\ast\}$, initial state $q_\varepsilon$, transitions $\delta(q_w, 0) = q_{w0}$ and $\delta(q_w, 1) = q_{w1}$ for all $w \in \{0, 1\}^\ast$ and accepting state set $F = \{q_w \mid w \in L\}$. This implies that even undecidable languages can be recognized by DIAs.
Yes, since we already know that $L = \{a^n b^n\}$ is not regular (using the pumping lemma), and that a language is regular if and only if it admits a DFA, it suffices indeed to show the inequivalency by definining a $DIA$ for $L$. Intuitively, $L$ is able to keep arbitrary amount of information about history using additional states when needed, and DFAs can keep only constant memory.
A way to achieve this $M=(\{a,b\}\times\mathbb{Z} \cup \{F\}, \delta, (a,0), \{(a,0),(b,0)\})$, where the first symbol in the tuple indicates whether a $b$ has been read so far, and the second counts $\#_a(x_0\dots x_i) - \#_b(x_0\dots x_i)$. So the transitions are $$\delta: ((a,i),a)\mapsto (a,i+1)\\ ((a,i),b)\mapsto (b, i-1)\\ ((b,i),b)\mapsto(b, i-1)\\ ((b,i),a)\mapsto F $$
