I'm having trouble understanding how there can be sufficiently many distinct elements for $\mathbb{R}$ to exist with its properties and yet still be a set.
(The --verbose explanation that follows includes all the bits that friends didn't understand immediately – just skip ahead once you got the idea of each section.)
1. sets are trees
Recall that sets are essentially just (weird) trees, and everything can be encoded as sets: $\newcommand{\set}[1]{\{#1\}}$ $\newcommand{\sO}{\emptyset}$ $\newcommand{\sSO}{\set{\sO}}$ $\newcommand{\sSSO}{\set{\sO,\sSO}}$ $\newcommand{\sSSSO}{\set{\sO,\sSO,\sSSO}}$ $\newcommand{\sSSSSO}{\set{\sO,\sSO,\sSSO,\sSSSO}}$
This tree/set could represent $\set{\sO,\set{\sSSO},\sSSSSO}$ or $\set{\sO,\set{2},\set{0,1,2,3}}$ or $\set{\sO,\set{(0,0)}, 4}$ or … – doesn't really matter. And the one on the left here…
…is really the same set as the one on the right, because the branches 2, 4, and 6 from the root are actually the same element. (They just look different because they are shown with "duplicate elements", which don't matter – look at the "triangle" at the end of the left set's rightmost branch: It depicts $\set{\sO,\sO}$, which is actually just $\set{\sO}$. You can keep dissecting stuff manually to identify the actual elements and get rid of the "ghosts"… but you don't need to:)
2. bounded size from bounded depth
There's an easier way to show that the example set above can't be that big: Look at the "depth" / rank of the tree – it limits the number of possible elements, and thereby the possible number of elements. Here's all sets of depth at most $d$:
(…and so on, giving at most 1, 2, 4, 16, 65536, 20[…19725 other digits…]36, … elements.)
The example set in the previous image has depth three, which means that all its elements have depth at most two, so there are at most 4 different elements available for forming the set. This means that the set on the left can't have the "claimed" size of 6, but can really contain at most 4 (distinct) elements. So we don't need to bother looking at the elements in detail to get an upper bound on the size, all we need to know is (a bound on) the depth.
3. the axiom of regularity / no infinite $\in$-chain
The axiom of regularity implies that no set has an infinite descending $\in$-chain (which is why $\in$-induction works). And so bounding the size of the set by the rank / maximum depth should work for all sets, right? That would give an upper bound of $2\uparrow\uparrow d$ elements for any set.
4. the set of real numbers
We also know that the real numbers are uncountable, i. e. the elements of $\mathbb{R}$ cannot be put into 1-to-1 correspondence with the elements of $\mathbb{N}$.
5. the problem
But if there cannot be infinite $\in$-chains, that means the depth is bounded, which bounds the number of possible distinct elements, which… keeps things countable? But then where are all those elements of $\mathbb{R}$ coming from?
What's the trick that makes this work without blowing up (i. e. no infinite $\in$-chain from which you could derive nonsense) while still preventing the collapse to countability?
A set does not have to have bounded depth! Consider the set $S=\{\emptyset,\{\emptyset\},\{\{\emptyset\}\},\{\{\{\emptyset\}\}\},\dots\}$. Each individual path down the tree of this set is finite, but there are infinitely many branches that split up at the first level and each has a different length, so there can still be infinitely many elements and unbounded depth.
Now note that we can take not just this set $S$ but any subset of it. Since $S$ is infinite, there are uncountably many different subsets of $S$. If we take the set $\mathcal{P}(S)$ of all subsets of $S$, we get an uncountable set. This set still can be represented by a tree with no infinite branches, it just branches a lot in the first two levels!