Would a square have more points than lines? Suppose you have a square. Now, let $L$ be the set of all lines that can be drawn in the area described by the square, and let $P$ be the set of all points in all the lines in $L$.
Would the cardinality of $P$ be greater or less than $L$, or is this one of those times in which the infinite desecrates our logic?
 A: They both have the same cardinality - namely, $2^{\aleph_0}$
(Note that contrary to what you may see in some popular mathematics textbooks, this is not the same thing as $\aleph_1$ - or rather, the question of whether $\aleph_1=2^{\aleph_0}$ is known to not be answerable from the usual axioms of set theory.)
In response to your comment

isn't a line defined by two points? How could the same amount of lines and points exist?

note that in fact every infinite set $A$ has the same cardinality as its square (= the set of pairs of elements of $A$)! This is an important difference between infinite and finite sets. Looking at Hilbert's hotel first might give you a good sense of how this can be the case.

A minor subtlety, which should be ignored until the rest of the answer is understood:
The statement "Every infinite set has the same cardinality as its square" is in fact equivalent to the axiom of choice. So there's a reasonable question here: if we disallow the axiom of choice, then can we distinguish between $\mathbb{R}$ and $\mathbb{R}^2$ in terms of cardinality?
However, the answer is still "no" - while the general fact requires choice, the specific instance we're interested in here is provable without choice. In fact, we can write down explicit bijections between $\mathbb{R}$ and $\mathbb{R}^2$.
In full detail, these are somewhat messy; however, let me give a hint which will hopefully let you come up with one yourself:


*

*First, let's try the simpler task of finding a bijection between $[0, 1)$ and $[0, 1)^2$. This certainly doesn't seem any easier.

*Well, any element of $[0, 1)$ is really just a sequence of digits - its decimal expansion. So I'm really asking how to turn two infinite sequences into one infinite sequence.

*There's a natural way to do this: interleaving! E.g. I can combine the sequences $111111111...$ and $345345345...$ to get the sequence $$131415131415131415...$$ So my first guess at a bijection is: "interleave the decimal digits."

*Turns out that doesn't quite work. (HINT: think about things like $0.99999...=1$.) Can you see how to fix it? Can you see how to then get a bijection from $\mathbb{R}$ to $\mathbb{R}^2$?
A: The cardinality of the whole square is $\mathfrak c\cdot\mathfrak c=2^{\aleph_0}\cdot 2^{\aleph_0}=2^{\aleph_0+\aleph_0}=2^{\aleph_0}=\mathfrak c $. So, $|P|=\mathfrak c $.
Now I wasn't sure if you meant straight lines or any (e.g. curved) lines, so I'll do the calculation for the latter. Each line in the square is a continuous map from $[0,1] $ to $P $, but it is uniquely determined by its values on $\mathbb Q\cap [0,1] $, so $|L|\le \mathfrak c^{\aleph_0}=(2^{\aleph_0})^{\aleph_0}=2^{\aleph_0\cdot\aleph_0}=2^{\aleph_0}=\mathfrak c $. Because, trivially, $|L|\ge\mathfrak c $, you end up with $|L|=\mathfrak c$.
Thus, $|P|=|L|=\mathfrak c $.
Finally, I was not sure, when you said "points in all the lines" whether you wanted the points in the square to be double-counted (as they belong to multiple lines). Still that number is at most $|L\times [0,1]|=\mathfrak c\cdot \mathfrak c=\mathfrak c $ - the same cardinality again!
