Show $C(\mathbb{R})$ is not separable.

These subsets are good to use. You only need to show the following general fact:

Let $(X, d)$ be a metric space, and $\Omega \subset X$ be an uncountable set so that $d(x, y) \ge 1$ for all $x, y\in \Omega$, $x\neq y$. Then $X$ is not separable.

To show this, assume the contrary that $X$ is separable. Let $\{ f_1, \cdots, f_n, \cdots, \}$ be a countable dense subset of $X$. Then

$$X = \bigcup_n B(f_i, 1/3).$$

As $\Omega$ is uncountable, there are $x, y\in \Omega$, $x\neq y$ so that $x, y\in B(f_i, 1/3)$ for some $i$. But that contradicts $d(x, y) \ge 1$.

Remark You might be also interested in the more "usual" proof: Let $\{f_1, \cdots, f_n, \cdots \}$ be a countable subset of $C(\mathbb R)$. Let $f \in C(\mathbb R)$ so that $\|f\| \le 1$ and $|f(n) - f_n(n)| \ge 1/2$ for all $n$. Then $\|f-f_n\| \ge 1/2$ for all $n$. Thus $\{f_1, \cdots, f_n, \cdots\}$ is not dense in $C(\mathbb R)$.