Mapping reduction from $A_{TM}$ to $INFINITE_{TM}$ same as to $ALL_{TM}$?
The m-reducibility definition says

Language $A$ is m-reducible to $B$ (denoted $A\leq_mB$) if there is a
computable function $f:\Sigma^* \rightarrow \Sigma^*$ such that for
every $w\in \Sigma^*$, $$w\in A \iff f(w)\in B$$

The point you should pay attention to is that there is no need to talk about every $x\in B$. If you can map all elements of $A$ to one special element of $B$ say $x_0$ and all strings that don't belong to $A$ to string which doesn't belong to $B$ say $y_0$ then you are done!
Maybe if you think of $A\leq_m B$ as language $B$ is harder than $A$ then you feel better about the reduction. This means if you can decide $B$ then you also can also decide $A$.
So by the proof, you have provided we map every $\langle M,w\rangle \in A_{TM}$ to $\Sigma^*$ and every $w' \notin A_{TM}$ to $\emptyset$. Thus $INFINITE_{TM}$ is harder than $A_{TM}$. If you could decide just one elements of $INIFINITE_{TM}$ which is $\langle M \rangle$ that $L(M)=\Sigma^*$ then you could decide $A_{TM}$.