Show $C(\mathbb{R})$ is not separable. The hint to do so is to show that $C(\mathbb{R})$ has an uncountable subset where the elements are very far apart. Here $C(\mathbb{R})$ has norm $$||f||_{C(\mathbb{R})} = \sup_{x \in \mathbb{R}}|f(x)|$$
My idea is as follows:  
Define the set $$\Lambda := \{f \in C(\mathbb{R}) : f(n) = n \, \text{or} \, \, f(n) = 0 \, \text{for each} \, n \in \mathbb{N}\}$$ Let $\Omega \subset \Lambda$ be the subset of piecewise linear functions. Then $\Omega$ is uncountable, since it is equivalent to the set of all binary sequences. Moreover, if $f,g \in \Omega$, then $$||f-g||_{C(\mathbb{R})} \geq 1.$$  
My other idea was to take $f(x) = 1-2|x|$ and shift it left and right, copy and paste it, and define the set of all possible shifts and pastes. In this case too, the difference between elements is $1$.
Say, functions like

Are either of these viable sets of functions to use? If so, I'm a bit confused as to how this implies the set is not countable. Is it because any countable subset has to be far from some of these functions?
 A: 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 $C(\mathbb R)$. 
A: The first set works (what you said is correct), and the second set works for the same reason. The collection of all linear combinations of the tent functions translated by integers contains the collection of tentlike- functions $f$ where $f(n) ∈ \{0,1\}$; this collection $\mathcal{C}$ is also in bijection with the space of binary sequences. 
Form the distinct radius 1/2 balls around your functions, $(B(f,1/2))_{f ∈ \mathcal{C}}$.  If $C(ℝ)$ were separable, then a countable dense subset $\{g_n\}$ would exist. The $g_n$s are covered by only countably many balls. The union of all the other balls would then constitute an open set whose points cannot be approximated by $g_n$s.
