I remember reading somewhere that the cardinality of a Banach space $X$, denoted $|X|$ is equal to that of the continuum, but couldn't find it.
If this is true, the set of all bounded functions from $\mathbb{R}$ to $\mathbb{R}$ with the uniform norm, which I'll denote with $\mathcal{B}(\mathbb{R},\mathbb{R})$ should not be a Banach space, since its cardinality is equal to $|\mathcal{P}(\mathbb{R})| > | \mathbb{R} |$. Is this true?
I think that showing that $\mathcal{B}(\mathbb{R},\mathbb{R})$ is a Banach space like this: It is clear that it is a normed space with the uniform norm since the functions are bounded. Now take a Cauchy sequence $(f_n)_{n \in \mathbb{N}} \subset \mathcal{B}(\mathbb{R},\mathbb{R})$. Then for all $\varepsilon > 0$ there exists a $N_{\varepsilon} \in \mathbb{N}$ such that $\| f_n - f_m \| < \varepsilon$. The only candidate for a limit $f$ of this Cauchy sequence is the pointwise limit $f(x) := \lim_{n \to \infty} f_n(x)$. How can I continue the proof, if the statement is correct?
No, the cardinality of a nontrivial Banach space is at least $\mathfrak{c}=\lvert\mathbb{R}\rvert$, but can be bigger. For example, let $S$ be a set of cardinality $> \mathfrak{c}$ equipped with the counting measure, then $\ell^1(S)=L^1(S)=\{f\colon S\to\mathbb{F}\mid \sum_{s\in S}\lvert f(s)\rvert=:\lVert f\rVert<\infty\}$ has cardinality at least $\lvert S\rvert>\mathfrak{c}$ given by coordinate indicator functions.
I think what you misremembered is
The proof is to use the countable dense subset $S$. There are $\lvert S^\mathbb{N}\rvert=\aleph_0^{\aleph_0}$ sequences from this countable dense subset, and each element of $X$ is represented by some Cauchy sequence from $S$, so $\lvert X\rvert\leq\aleph_0^{\aleph_0}=\mathfrak{c}$. Since $X$ needs to contain a copy of $\mathbb{R}v$ for some nonzero $v\in X$, we have $\lvert X\rvert=\mathfrak{c}$.