I think the answer is no, as shown by the following set $A$ (whose elements I will describe as subsets rather than binary functions).
For any map $\sigma:\omega\to\omega+1$ \sigma:\omega\to\omega+1:=\omega\cup\{\omega\}$define its hypograph $$\mathrm{hypo}(\sigma):=\{(m,k)\in\omega\times\omega: k < \sigma(m)\}\subset\omega\times\omega\ .$$ An increasing map$\sigma:\omega\to\omega+1$takes the value$\omega$if and only if its hypograph contains a subset of the form$[n,\omega)\times \omega$, thus, if and only if it has the property (P) stated in the question (there exist infinitely many$m$such that there exist infinitely many$k$such that$(m,k)\in \mathrm{hypo}(\sigma)$; which is indeed verified quite in a strong sense)way). Let$A$be the set of all hypographs of all increasing maps that take the value$\omega$. Clearly, the set$A$is countable, with no isolated points; and all its elements enjoy property (P). A point in$\mathrm{cl}(A)\setminus A$is exactly the hypograph of an increasing$\omega$-valued map, which never satisfy satisfies property (P). 1 I think the answer is no, as shown by the following set$A$(whose elements I will describe as subsets rather than binary functions). For any map$\sigma:\omega\to\omega+1$define its hypograph $$\mathrm{hypo}(\sigma):=\{(m,k)\in\omega\times\omega: k < \sigma(m)\}\subset\omega\times\omega\ .$$ An increasing map$\sigma:\omega\to\omega+1$takes the value$\omega$if and only if its hypograph contains a subset of the form$[n,\omega)\times \omega$, thus, if and only if it has the property (P) stated in the question (there exist infinitely many$m$such that there exist infinitely many$k$such that$(m,k)\in \mathrm{hypo}(\sigma)$; which is indeed verified quite in a strong sense). Let$A$be the set of all hypographs of all increasing maps that take the value$\omega$. Clearly, the set$A$is countable, with no isolated points; and all its elements enjoy property (P). A point in$\mathrm{cl}(A)\setminus A$is exactly the hypograph of an increasing$\omega\$-valued map, which never satisfy property (P).