I'm once again trying to define a meaningful notion of size for sets.
Let $R$ be a set of size $\mathfrak c=2^{\aleph_0}$ and say $\mathcal F\subset \mathcal P(R)$ is a family of humongous sets if it has the two following properties:
- If $P = \{P_n: n\in\mathbb N\}$ is a countable partition of $R$, then $P\cap\mathcal F\ne\varnothing$, i.e., there is an humongous set in $P$.
- If $\{H_n: n\in\mathbb N\}\subset\mathcal F$ is a countable family of humongous sets, then $\left|\displaystyle\bigcap_{n\in\mathbb N} H_n\right| = \mathfrak c$.
Now I wonder, of course, if one can find a family of humongous sets.
While wondering I noticed the chosen cardinals are not really doing anything for the definition, so let's generalize it!
Let $S$ be a set, $\alpha$ a cardinal and say $\mathcal F\subset \mathcal P(S)$ is a family of $\alpha-$humongous sets if it has the two following propeties:
- If $P=\{P_\kappa: \kappa<\alpha\}$ is a partition of $S$, then $P\cap \mathcal F\ne\varnothing$.
- If $\{H_\kappa: \kappa<\alpha\}\subset\mathcal F$, then $\displaystyle\left|\bigcap_{\kappa<\alpha} H_\kappa\right| = |S|$.
If the answear to my first question is false, one could ask if there is some size for $S$ so that it admits a family of $\aleph_0-$humongous sets and, if there is, what is the smallest such size.
At least taking $\vert S\vert=\alpha$ you're basically looking for the notion of a measurable cardinal: $\kappa$ is measurable iff its powerset admits a family of sets $\mathcal{F}$ satisfying your conditions which is also closed upwards (which doesn't add much to what you've got - in particular, the upwards closure of any family satisfying your properties still satisfies your properties).
Actually, since you're mainly interested in the $\alpha=\aleph_0$ case, you're really looking for Ulam measurable cardinals. But the smallest Ulam measurable, if it exists, is also the smallest measurable, so at the level of "How big are such things?" the notions are no different.
Here's a proof that your first question has a negative answer. Without loss of generality, assume $R$ is the set of real numbers and suppose $\mathcal{F}$ had the desired properties. By your first bulletpoint there is some integer $z$ such that $[z,z+1)\in\mathcal{F}$; consider the partition of $R$ into intervals of this form.
Now refine this partition by replacing the single block $[z,z+1)$ with two blocks $[z, z+{1\over 2})$ and $[z+{1\over 2}, z+1)$. By points 1 and 2, one of these two pieces must be in $\mathcal{F}$. Iterating this strategy, we can find a sequence of intervals each of which is in $\mathcal{F}$ but such that the diameter of the $n$th interval is $2^{-n}$. Now apply your second bulletpoint to this sequence. The intersection of these intervals has at most one element, so we get a contradiction.
It is not immediate that measurable cardinals must be particularly big, but in fact this is the case; indeed, the usual axioms of set theory cannot prove that measurable cardinals exist, and the smallest measurable cardinal (if it exists) is truly gigantic. See e.g. 1, 2, although these posts are necessarily a bit technical.