It depends on whether you are talking about finite sequences, infinite sequences, or "uncountable sequences". Here is an attempt to give you a little bit of an intuition for the mathematical territory:

If you are talking about the set of all finite real sequences, then we have the following argument: for any $n$, the cardinality of $\mathbb{R}^n$ is the same as the cardinality of $\mathbb{R}$ (which I will call $c$ for convenience). Thus, the set of finite sequences of a given length is a set of cardinality $c$. The cardinality of the arbitrary union of sets of cardinality $c$ gives you another set of cardinality $c$. Thus, the set of all finite sequences is of cardinality $c$. The same argument can be made regarding $\mathbb{Q}$ (which has cardinality $\aleph_0$)

The cardinality of infinite sequences is, however, a different story. Cantor's argument tells us that for any set $S$, we have $\left| \wp(S) \right|>\left| S \right|$. For every subset of the rational numbers, there is a corresponding sequence in $\mathbb{Q}^{\mathbb{N}}$. Real numbers, on the other hand, are uncountable, so no infinite sequence will contain every element. So, as it ends up,
$\left| \mathbb{R}^{\mathbb{N}} \right| =\left| \mathbb{Q}^{\mathbb{N}} \right|= c$

Finally, we have the "uncountable sequences", that is, the set of functions from $\mathbb{R}$ to $\mathbb{R}$, which does have a greater cardinality. Here, we can use the previous argument as follows: for any subset $S \subseteq \mathbb{R}$, we can map $S$ to a function $f(x)$ for which $f(x)=0$ for any $x\in S$. This gives us an injective map from $\wp(\mathbb{R})$ to the set of functions from $\mathbb{R}$ to itself. Thus, the cardinality of the set of functions from $\mathbb{R}$ to $\mathbb{R}$ is greater than $c$.

A nice bit of cardinal arithmetic to have in your arsenal: for transfinite sets $P$ and $Q$:
$$
\left| P^Q\right|=\max\{\left|P\right|,\left|\wp(Q)\right|\}
$$
EDIT: I am not sure about that last formula, let's call it a conjecture.