Let $A=\{f\mid f:(0,1)\rightarrow\mathbb{R}\mbox{ is continuous}\}.$
Let $B=\mathbb{R}^{\mathbb{Q}\cap(0,1)}=\{g\mid g:\mathbb{Q}\cap(0,1)\rightarrow\mathbb{R}\}$
be the set of all real-valued functions defined on $\mathbb{Q}\cap(0,1)$.
Clearly $|B|=c^{\omega}=(2^{\omega})^{\omega}=2^{\omega\times\omega}=c$.
Define $\theta:A\rightarrow B$ by $\theta(f)=f|_{\mathbb{Q}\cap(0,1)}$
(i.e., $f$ restricted on $\mathbb{Q}\cap(0,1)$). Note that $\theta$
is injective. For, let $f_{1},f_{2}\in A$ such that $\theta(f_{1})=\theta(f_{2})$.
Let $x\in(0,1)$ be arbitrary. Choose a sequence $(r_{n})$ in $\mathbb{Q}\cap(0,1)$
such that $r_{n}\rightarrow x$. By continuity, we have
\begin{eqnarray*}
f_{1}(x) & = & \lim_{n}f_{1}(r_{n})\\
& = & \lim_{n}\theta(f_{1})(r_{n})\\
& = & \lim_{n}\theta(f_{2})(r_{n})\\
& = & \lim_{n}f_{2}(r_{n})\\
& = & f_{2}(x).
\end{eqnarray*}
Therefore, $f_{1}=f_{2}$.
It follows that $|A|\leq|B|=c$. For each $a\in\mathbb{R}$, define
$\bar{a}:(0,1)\rightarrow\mathbb{R}$ by $\bar{a}(x)=a$. (i.e., $\bar{a}$
is a constant function). Let $A_{0}=\{\bar{a}\mid a\in\mathbb{R}\}$.
Clearly, the map $\mathbb{R}\rightarrow A_{0}$, $a\mapsto\bar{a}$
is injective, so $c=|\mathbb{R}|\leq|A_{0}|$. Obviously, $A_{0}\subseteq A$.
Therefore, we have $c\leq|A_{0}|\leq|A|\leq c$. This shows that $|A|=c$.
