Prove the existence of some weird family of sets

Here's a construction which I think combines the cleanest elements of Noah's and Gro-Tsen's.

By applying bijections, we can use any sets $A,B$ with $|A| = |\mathbb{N}|$ and $|B| = |\mathbb{R}|$ in place of $\mathbb{N}$ and $\mathbb{R}$.

Consider the infinite rooted binary tree. Take $A$ to be the set of $k$-tuples of vertices in the tree, where we require the vertices in each $k$-tuple to all lie on the same level of the tree. Take $B$ to be the set of infinite rays in the tree which start at the root. These sets have the correct cardinalities.

For each ray $\rho$ in $B$, let $X_\rho$ be the set of $k$-tuples in $A$ which contain at least one element of $\rho$.

The intersection of any $k$ sets of this form is infinite, since on each level of the tree we can choose a $k$-tuple containing one vertex from each ray.

The intersection of any $k+1$ sets of this form is finite, since on some level of the tree, the $k+1$ rays must have all branched apart, so there is no $k$-tuple of vertices on that level or below which hits all $k+1$ rays.

For each $\xi\colon \mathbb{N}\to\{0,1\}$, consider the set $X_\xi$ of functions $s\colon \{0,\ldots,N-1\} \to\{0,1\}$ where $N\in\mathbb{N}$ and such that there exists $j\in\{0,\ldots,k-1\}$ for which $s(ik+j)=\xi(i)$ for each $i$ such that $ik+j<N$ (roughly: the values of $s$ on some congruence class mod $k$ must be those given by $\xi$).

Given $k$ functions $\xi_j\colon \mathbb{N}\to\{0,1\}$ (for $0\leq j<k$), we can find infinitely many $s$ in the intersection of the $X_\xi$, namely the retriction $\{0,\ldots,N-1\}$ for any $N$ of the function $ik+j \mapsto \xi_i(j)$. Given at least $k+1$ of them, however, any way of assigning congruence classes mod $k$ to them will have two different $\xi$'s in the same class, so that there are only finitely many $N$'s up to which one can construct a function $s$ agreeing with the given $\xi$'s on those congruence classes.

This constructs for each $\xi \in \{0,1\}^{\mathbb{N}}$ a set $X_\xi$ of finite binary sequences with the required intersection properties (any $\leq k$ of them have infinite intersection, any $\geq k+1$ of them have finite intersection). Now it's just a matter of injecting the reals in $\{0,1\}^{\mathbb{N}}$ and injecting the set $\{0,1\}^*$ of finite binary sequences in the natural numbers.

(Reading Noah Schweber's solution, this one appears to be essentially equivalent, only using finite sequence approximations to infinite sequences instead of rational approximations to reals. I wonder if very different constructions can be found.)

OK, here's a really silly way to do it:

A $k$-Cauchy sequence is a sequence of $k$-tuples $((x_1^1, . . . , x_k^1), (x_1^2, . . . , x_k^2), . . .)$ which is componentwise Cauchy - that is, for each $i$ the sequence $(x^1_i, x^2_i, x^3_i, . . . )$ is Cauchy. (If you think of a Cauchy sequence as approximating a real, think of a $k$-Cauchy sequence as approximating a $k$-tuple of reals.)

A $k$-Cauchy sequence of rationals is just a $k$-Cauchy sequence, all of whose terms are $k$-tuples of rationals.

OK, fix a bijection $f: \mathbb{Q}\cong \mathbb{N}$; and for each real $r$, fix a specific Cauchy sequence of rationals $C_r$ converging to $r$. Note that from this we can associate a $k$-Cauchy sequence of rationals $C_F$ to any $k$-tuple of reals $F$.

Now via $f$ we can identify initial segments of $k$-Cauchy sequences of rationals with natural numbers. (It's a bit messy, but oh well.) To the real $r$ we assign the set $A_r$ of all naturals associated to initial segments of $k$-Cauchy sequences of rationals $C_F$ with $F\ni r$, that is, $$A_r=\{code(\sigma): \sigma\prec C_F\mbox{ for some $F\ni r$ with $\vert F\vert=k$}\}$$ where "$code(\sigma)$" is the natural associated to $\sigma$.

Whew! Well, now, let's see that this satisfies the desired properties. The intersection of $k$-many $A_r$s is clearly infinite; the hard part is showing that the intersection of $(k+1)$-many $A_r$s is finite. This is a neat argument, which for now I'm going to leave as an exercise, with a hint:

HINT: Suppose $A_{r_1}\cap . . . \cap A_{r_{k+1}}$ is infinite. Then show that for some $i\le k$, infinitely many initial segments of the Cauchy sequences $C_{r_i}$ and $C_{r_{k+1}}$ are the same. What does this tell you about $r_i$ and $r_{k+1}$?

FURTHER HINT: To find the $i$ mentioned in the previous hint, look at $C_{(r_1, . . . , r_k)}$. Since the $r_i$s are disjoint, there's some $n$ such that $C_{r_i}(n)\not=C_{r_j}(n)$ for any $i<j\le k$ (why?). So look at the $n$th term of $C_{(r_1, . . . , r_k)}$. This is a $k$-tuple, and one of the elements of this $k$-tuple must be the length-$n$ initial segment of $C_{r_{k+1}}$ (why?) . . .