Prove a set of continuous function is bounded. Let $C_b(\mathbb{R})$ denote the space of all bounded, continuous functions $f : \mathbb{R} → \Bbb C$. Let $C_0(\mathbb{R})$
denote the set of continuous functions $f : \mathbb{R} \to \Bbb C$ for which
lim $f(x) = 0$ as $x\to\pm \infty$.
a) Prove that every $f ∈ C_0(\mathbb{R})$ is bounded.
So I need to show $|f| < M$ for some $M$.
b) Prove that $C_0(R)$ is closed in $C_b(R)$ 
I just really need help.  I don't know how to show these two things. 
I know that $d$ makes $C_b(\mathbb{R})$ into a complete metric space.  That's the only Theorem from class I really have.  
 A: Since $f$ tends to zero as $|x|\to \infty$, there exist $R_0$ such that $|f(x)|\leq 1$ for every $|x|\geq R_0$. Due to compactness of $[-R_0,R_0]$ we find that $f$ is bounded there by a constant $M$. Thus $\|f\|_\infty\leq \max\{M,1\}$
Let $f_k \in C_0(\mathbb{R})$ be given and such that $f_k\to f$ uniformly (i.e. in $C_b(\mathbb{R})$. This is an assumption! If $f$ turns out to be in $C_0$, we have shown that $C_0$ is closed in $C_b$ by definition. We need to show that $f(x)=0$ as $|x|\to\infty$.
For every large enough $|x|$ and some fixed $k$, we find the estimate $$|f(x)|\leq |f(x)-f_k(x)| +|f_k(x)| \leq  \varepsilon$$
The first term in the middle is small due to the uniform convergence, the second term tends to zero for fixed $k$. 
@ OP: Can you give the precise argument here?
A: Problem a) is easy.
Ad b): I'm assuming that your distance $d$ is defined by $d(f,g):=\sup_{-\infty<x<\infty}|f(x)-g(x)|$. Consider a function $f_0\in C_b\setminus C_0$. Then there is an $\epsilon>0$ and a sequence $(x_n)_{n\geq1}$ with $|x_n|\to\infty$ $\>(n\to\infty)$ and $|f_0(x_n)|\geq 2\epsilon$ for all $n\geq1$. 
Now the set 
$$\bigl\{g\in C_b\>\bigm|\>d(g,f_0)<\epsilon\bigr\}$$
is a neighborhood of $f_0$ that does not intersect $C_0$.
