How to reduce halting problem to the problem of whether a Turing Machine accepts infinitely many inputs?
If it is decidable, let $M$ decide it. 

Construct a decider $D$ that works on input $\langle w_1,w_2\rangle$ as follows:


Construct $M_w$ (as well as its encoding $w$) that works on input $x$ as follows:


Run $M_{w_1}$ on $w_2$.
Accept.

Run $M$ on $w$.
If $M$ accepts, accept; otherwise, reject.


We can see $D$ accepts if and only if $M$ accepts $w$, i.e. $M_w$ accepts infinitely many inputs, which means $M_{w_1}$ halts on $w_2$. Therefore $D$ is a decider for the halting problem, a  contradiction.

Hence the language given in OP is undecidable.