Proof by Turing Reduction
Your solution doesn't quite make sense for many reasons:


The reduction should take $\langle M \rangle$ as input, and using an oracle to $Q$, decide whether $\langle M \rangle \in I$.
The reduction should be deterministic (so step 3 is invalid).
The function of the arbitrary word $u$ is not explained; indeed, steps 1 and 2 seem useless.
You don't explain what $M_1,M_2$ are in step 4 (are you trying to reduce from $Q$ to $I$?).
You don't explain how to implement step 4.


Also, you are missing a proof that your reduction works. Without a proof, your solution is wrong even if it could be completed to a correct solution.

One solution goes along the following lines (I'm assuming $L(M)$ consists of all inputs on which $M$ halts; I also say that $n \in L(M)$ is accepted by $M$):


Given $M$, let $M'$ be the machine that on input $n$, simulates $M$ on $n$, and then (if $M$ halted) tries to find an $m &gt; n$ accepted by $M$ (if there is no such $M$, it never halts); you can implement this using the technique of dovetailing.
Give $M$ and $M'$ to the $Q$ oracle, and return its answer.


The idea is that $L(M)$ is infinite iff $L(M) = L(M')$. I'll leave you the details.