From Jech's Set Theory, we have the following theorems on cardinal exponentiation (a Corollary on page 49):
Theorem. For all $\alpha,\beta$, the value of $\aleph_{\alpha}^{\aleph_{\beta}}$ is always either:
- $2^{\aleph_{\beta}}$; or
- $\aleph_{\alpha}$; or
- $\aleph_{\gamma}^{\mathrm{cf}\;\aleph_{\gamma}}$ for some $\gamma\leq\alpha$ where $\aleph_{\gamma}$ is such that $\mathrm{cf}\;\aleph_{\gamma}\leq\aleph_{\beta}\lt\aleph_{\gamma}$.
Here, $\mathrm{cf}\;\aleph_{\gamma}$ is the cofinality of $\aleph_{\gamma}$: the cofinality of a cardinal $\kappa$ (or of any limit ordinal) is the least limit ordinal $\delta$ such that there is an increasing $\delta$-sequence $\langle \alpha_{\zeta}\mid \zeta\lt\delta\rangle$ with $\lim\limits_{\zeta\to\delta} = \kappa$. The cofinality is always a cardinal, so it makes sense to understand the operations above as cardinal operations.
Corollary. If the Generalized Continuum Hypothesis holds, then
$$\aleph_{\alpha}^{\aleph_{\beta}} = \left\{\begin{array}{lcl}
\aleph_{\alpha} &\quad & \mbox{if $\aleph_{\beta}\lt\mathrm{cf}\;\aleph_{\alpha}$;}\\
\aleph_{\alpha+1} &&\mbox{if $\mathrm{cf}\;\aleph_{\alpha}\leq\aleph_{\beta}\leq\aleph_{\alpha}$;}\\
\aleph_{\beta+1} &&\mbox{if $\aleph_{\alpha}\leq\aleph_{\beta}$.}
\end{array}\right.$$
So, under GCH, for all cardinals $\kappa$ with cofinality greater than $\aleph_0$ have $\kappa^{\aleph_0} = \kappa$, and for cardinals $\kappa$ with cofinality $\aleph_0$ (e.g., $\aleph_0$, $\aleph_{\omega}$), we have $\kappa^{\aleph_0} = 2^{\kappa}$. (In particular, it is not the case the cardinality of $A^{\mathbb{N}}$ is necessarily less than the cardinality of $\mathcal{P}(A)$).
Then again, GCH is usually considered "boring" by set theorists, from what I understand.
(Note that $|\Bbb C|=|\Bbb R|$.)
If we assume the axiom of choice, then indeed the answer is that the set of countable subsets of $\Bbb R$ has the same cardinality as $\Bbb R$ itself.
The proof is simple. Using the axiom of choice, choose an enumeration of each such subset (i.e. a bijection between the countable set and $\Bbb N$). Next note the following cardinal arithmetic:
$$|\Bbb{R^N}|=(2^{\aleph_0})^{\aleph_0}=2^{\aleph_0\cdot\aleph_0}=2^{\aleph_0}=|\Bbb R|.$$
Therefore the set of sequences with real coefficients have the same cardinality as the real numbers. Now send each countable set to its enumeration, and we're done.
Without the axiom of choice, however, it is consistent that there are strictly more countable subsets. For example in Solovay's model where every set is Lebesgue measurable this is true.
Best Answer
$\mathbb R[X]$ has the same cardinality as $\mathbb R$ itself.
One fairly simple way to see this is to know that there are bijections $f: \mathbb R \to \mathcal P(\mathbb N)$ and $g: \mathbb N\times\mathbb N \to \mathbb N$. Then
$$h(a_0+a_1X+\cdots a_n X^n) = \{g(p,q)\mid p\in f(a_q)\}$$ defines an injection $h:\mathbb R[X]\to \mathcal P(\mathbb N)$, and since there are obviously at least as many polynomials as there are real numbers, the Cantor-Bernstein theorem takes care of the rest.