Before proving it, let's resolve an ambiguity in the statement of the problem. The statement "$n$ is the product of at most $k$ prime numbers" must be taken to mean "$n$ can be written as the product of at most $k$ prime numbers, not necessarily distinct." (If we took it to mean "$n$ has at most $k$ prime factors", then $A=$ the set of all powers of $2$ would be a counterexample.)
It was pointed out in a comment (apparently now deleted) that the problem asks for a proper subset of $A.$ In fact, it's sufficient to show that for any set $A$ with the given properties, we can find an infinite $B \subseteq A$ such that any two members of $B$ have the same gcd. After all, if there is such a $B,$ then $B$ with its first element removed is a proper subset of $A$ with the required property.
We'll prove the statement by induction on $k,$ as the OP suggests.
For $k=1,$ take $B=A.$
Now assume the statement is true for $k,$ and let $A$ be an infinite set of positive integers such that each $n \in A$ is the product of at most $k+1$ primes.
Case 1: There exists some number $m > 1$ such that infinitely many multiples of $m$ belong to $A.$ Set $A' = \lbrace n \;\vert\; mn \in A \rbrace,$ and observe that $A'$ is infinite and each member of $A'$ is the product of at most $k$ primes. Let $B$ be an infinite set contained in $A'$ such that the gcd of any two distinct members of $B$ is some fixed number $g.$ Then $\lbrace mb \;\vert\; b \in B \rbrace$ is an infinite subset of $A$ such that any two distinct members of $B$ have gcd $mg.$
Case 2: For every $m,$ only finitely many multiples of $m$ belong to $A.$ Define $b_0, b_1, \dots \in A$ by induction, as follows: let $b_n$ be the least member of $A$ that is greater than 1 and that is not a multiple of any prime that divides some $b_k$ for $k<n.$ Then $B = \lbrace b_0, b_1, \dots \rbrace$ is an infinite subset of $A$ such that any two distinct members of $B$ have gcd $1.$