Is the halting problem decidable by an "infinite Turing machine"? It has been shown of course that the halting problem is undecidable. That is, we cannot formulate a Turing machine that will decide for any arbitrary turing machine whether it will halt or not. 

However, I'm wondering, if we want to decide for the set of Turing machines $T$, whether a turing machine $t$ halts, and we somehow have access to "infinite Turing machines" $I$, which have the property $P$: "They can calculate an infinite amount of steps in finite time", would it then be possible to grab a $i$ from $I$ that decides the halting problem for all $t\in T$? 

The proof of the undecidability of the halting problem no longer applies here, because it would require one of the $t\in T$ to contain $i$, but then we might possibly have that $t\in I$ (since $t$ now simulates $i$, which is in $I$). 

So my question is, can we prove that in this case, the halting problem is decidable? If not, are there other constructions of $I$ than the one I came up with that would make it possible?

(is there a name for what I've called "infinite turing machines"?)
It is true that infinite Turing Machines as you have defined them can solve the halting problem for regular Turing Machines. That is, there exists an infinite TM, $I(t,x)$, that returns $1$ iff $t(x)$ halts. To see this, just have your first step be “run t(x) for infinitely many steps” and the rest is easy after that. Yes, there are some issues formalizing this idea, but it can be done. This is called a “Hyper Turing machine”

There’s an interesting question lurking here though... is there an infinite TM that solves the halting problem for infinite Turing Machines? The answer to that is no, and if you know the diagonalization proof of the incomputability of the halting problem that applies here too.