The set of bounded continuous functions is a closed subspace of that of bounded functions suppose S is a metric space and $B(S)$ is the set of bounded functions and $C_b(S)$ is the set consisting of bounded continuous functions.
Prove that $C_b(S)$ is a closed subspace of $B(S)$.
I thought of looking at the complement $B(S) \backslash C_b(S) = \{f| \text{ f is bounded and not continuous}\}$ and proving this is open. 
But how does one check that for each element $f \in  B(S) \backslash C_b(S)$, there exists an $\epsilon$ such that $B_f(\epsilon) \subset B(S) \backslash C_b(S)$?
 A: The norm on the function space is given by the sup norm $\|f\|:=\sup\{|f(x)|x\in S\}$. Convergence of functions in sup-norm is called uniform convergence. It is a general statement that a uniform limit of continuous functions is again continuous.
To see that let $x_n \to x$, let $f_m \in C_b(S)$ and $f_m \to f \in C(s)$. Consider $\epsilon>0$.
Since $f_m \to f$ there exists an $M$ st for $m>M$, $\|f_m-f\|<\epsilon$. Now let $m>M$, since $f_m$ is continuous and $x_n\to x$ we have an $N$ so that $|f_m(x_n)-f_m(x)|<\epsilon$. So for a given $\epsilon$, choose $m, N$ as above and if $n>N$:
$$|f(x_n)-f(x)|≤|f(x_n)-f_m(x_n)|+|f_m(x_n)-f_m(x)|+|f_m(x)-f(x)|≤2\|f-f_m\|+\epsilon≤3\epsilon$$
Which shows $f(x_n)\to f(x)$ and $f$ is continuous.
A: An alternative answer (based more on OP's knowledge):
In a metric space $(X, \Vert \cdot \Vert_X)$ and $A \subset X$ we have the following assertion:
$$ x \in \bar{A} \iff \exists (x_n)_{n \in \mathbb N} \subset A: x_n\to x \in X $$
This means in words: an element $x$ of a subspace $A \subset X$ is in the closure of $A$ (the smalles closed set containing $A$) if and only if there is a sequence in A that converges in X. To see that this is equivalent to the definition of closed I leave for you to check with 2 hints:
First: Use the toplogical definition of a closure.
Second: How does a topology induces by a metric look like? How is convergence defined?
Now with this at hands we pick an arbitrary sequence $f_n$ in $C_b(S)$ which does convergence in $B(S)$ ! If the limit, say $f$, now lies in $C_b(S)$ we have shown that $C_b(S)=cl(C_b(S))$. To conclude see post above.
