The real line has cardinality at most $\aleph_2$, but transfinite ordinal space has arbitrarily high cardinality: what is wrong? In the context of supertasks, people and mathematicians are comfortable with the idea of transfinite ordinal time, that is, that time can be divided into an arbitrarily high number of steps. In most cases the number of steps is limited to be $\omega$, but some models, such as Hamkins infinite time Turing machines, assume that a finite amount of time  can be divided into a number of steps of arbitrarily high cardinality. I think we can safely extend the concept from time to space (actually the question is the same, just that I think many people will find it easier to identify space with the real line).  Then, the original question: The real line has $2^{\aleph_0}$ (which I guess is at most  $\aleph_2$) points. But if we can partition it into a number of intervals of arbitrarily high cardinality, shouldn't the number (or set) of points on it have at least the same cardinality? (or you can have more intervals than points?). I am obviously confused. Please help!! 
 A: First of all, the real line can be of size $\aleph_2$, but also of size $\aleph_{5223435}$ and even $\aleph_{\omega_1}$. All these are consistent with ZFC, and unless you assume something additional you can't really prove a lot about the cardinality of the continuum.
Secondly, you are correct. Assuming the axiom of choice a set cannot be partitioned into a strictly larger number of parts. That means that if $\Bbb R$ has cardinality $\aleph_2$, then every partition must have size of at most $\aleph_2$. However, do note that there are $\aleph_3$ many ordinals of cardinality $\aleph_2$.
A: $2^{\aleph_0}$ can be almost anything, it is not limited to $\aleph_2$.  As long as you can't divide the line into more than $2^{\aleph_0}$ segments, the fact that $|2^{\aleph_0} \times 2^{\aleph_0}|=|2^{\aleph_0}|$ means you have no trouble with more points than intervals.  Whatever $2^{\aleph_0}$ is, $\mathbb R^n$ has that number of points in it, too.
A: Hamkins's construction doesn't really assume "assume that a finite amount of time can be divided into a number of steps of arbitrarily high cardinality". Is merely proposes using an arbitrary ordinal (of whatever cardinality) as the time coordinate for the Turing machine computation, and investigates the consequences of such a decision. It doesn't depend on the full ordinal indexed time axis to correspond to "a finite amount of time", or to a subset of $\mathbb R$.
Indeed, Theorem 1.1 of the article you link to says that even if we don't assume a particular cardinality of the time axis, it is impossible for the machine to halt after more than countable steps. So essentially, the possibility of an uncountably long computation is allowed by the definition only to permit an argument that it is not an interesting case; all we really need to consider is computations that terminate in less than $\omega_1$ time.
Now it is well known that every ordinal below $\omega_1$ can be embedded not only into the real interval $[0,1]$, but can even be embedded into the rationals between $0$ and $1$. On the other hand, $\omega_1$ itself is not order-isomorphic to any subset of $\mathbb R$.
On yet another hand, that may not matter (at least if we restrict our attention to finite input tapes -- which, however, is a pretty big if), because there are only countably many different Turing machines, so since $\omega_1$ is a regular cardinal, there will be some countable ordinal before which every terminating computation has terminated. And that upper bound can be embedded into $\mathbb Q\cap[0,1]$.

As a philosophical comment to your question, the real line is merely a (fairly good) mathematical model of physical time. It may or may not correspond to actual physical time, and there seems to be no particular reason to insist that the hypothetical, non-physical, "philosophical time" that the "supertask" concept evokes ought to be constrained to things that can be modeled by the real line. Why not the long line, for example?
