I have come across this exercise, and I already know of a proof that proves that the cardinality of the set of all functions $\mathbb{R} \rightarrow \mathbb{R}$ is bigger than the cardinality of $\mathbb{R}$. Here's what I would do:
Suppose there exists a bijection $\Phi$ that maps $\mathbb{R}$ to $F$, so that $\Phi(x) = f_x$, so that all functions from $R$ to $\{0, 1 \}$ are contained among the functions $f_x, x \in \mathbb{R}$. Now, let's define $g : \mathbb{R} \rightarrow \{ 0, 1 \}$ so that $g(x) \ne f_x(x)$. It's clear that such a function can not be contained among the functions $f_x, x \in \mathbb{R}$ because its value differs from all functions $f_x, x \in \mathbb{R}$, for at least one value of $x$.
My question is, is this a sufficient proof that, at least, the cardinality of $F$ is different from the cardinality of $\mathbb{R}$?
Your prove is correct just need to rephrase, you should be clear how to construct $g(x)$. Suppose there is a bijection between $F$ and $\mathbb{R}$, $\Phi(x): \mathbb{R}\to F$ let $\Phi(x) = f_x$ we will construct $g: \mathbb{R}\to \{0,1\}$ different from $f_x, x\in \mathbb{R}$. Define $$g(x) = \begin{cases} 0 & f_x(x)=1 \\ 1 & f_x(x)=0\end{cases}$$ Clearly $g(x)$ is non of the $f_x$ for any $x\in \mathbb{R}$ since $g(x)\ne f_x(x)$ a contradiction $\Phi(x)$ is not a bijection.