The cardinality is at least that of the continuum because every real number corresponds to a constant function. The cardinality is at most that of the continuum because the set of real continuous functions injects into the sequence space $\mathbb R^N$ by mapping each continuous function to its values on all the rational points. Since the rational points are dense, this determines the function.
The Schroeder-Bernstein theorem now implies the cardinality is precisely that of the continuum.
Note that then the set of sequences of reals is also of the same cardinality as the reals. This is because if we have a sequence of binary representations $.a_1a_2..., .b_1b_2..., .c_1c_2...$, we can splice them together via $.a_1 b_1 a_2 c_1 b_2 a_3...$ so that a sequence of reals can be encoded by one real number.
Fix a strictly increasing sequence whose limit is $1$, for example: $$a_n=\frac{n}{n+1}$$
Now for every $c\in (0,1)$ the sequence $c_n=c\cdot a_n$ is strictly increasing and converges to $c$. Thus we easily obtain that there are at least $2^{\aleph_0}$ many strictly increasing sequences.
On the other hand every sequence is a function from $\mathbb N$ into $\mathbb R$. Since $\left|\mathbb R\right|=\left|\{0,1\}^\mathbb N\right|$ we have that $\left|\mathbb R^\mathbb N\right|=\left|\left(\{0,1\}^\mathbb N\right)^\mathbb N\right|$.
From How to show $(a^b)^c=a^{bc}$ for arbitrary cardinal numbers? we deduce that this set, therefore, has the same cardinality as $\{0,1\}^{\mathbb N\times\mathbb N}$. Using Cantor's pairing function (or a different method) we have that $|\mathbb N|=|\mathbb N\times\mathbb N|$, thus obtaining:
$$\left|\mathbb R^\mathbb N\right|=\left|\{0,1\}^{\mathbb N\times\mathbb N}\right|=\left|\{0,1\}^\mathbb N\right|=|\mathbb R|$$
So we have that the set of all strictly increasing sequences is a subset of $\mathbb R^\mathbb N$, so it cannot be more than $2^{\aleph_0}$ but on the other hand we have at least $2^{\aleph_0}$ many sequences in this set.
By Cantor-Bernstein we therefore have that this set has $2^{\aleph_0}$ many elements.
Best Answer
HINT: Use the fact that $|A^B|=|A|^{|B|}$, and show that $|\Bbb{R^R}|=2^{|\Bbb R|}$.