Proof that the uniform convergence topology is finer than the pointwise convergence topology Let $C(\mathbb{R})$ be the set of continuous functions over $\mathbb{R}$. I already showed that the collection of all following sets are a basis for the uniform convergence topology:
For $f \in C(\mathbb{R})$ and $\varepsilon > 0$ 
\begin{equation*}
V(f, \varepsilon) = \left\{ g \in C(\mathbb{R}) \mid \sup_{t \in \mathbb{R}} |g(t) - f(t)| < \varepsilon  \right\}
\end{equation*}
And that the collection of all following sets are a basis for the pointwise convergence topology:  for $f \in C(\mathbb{R})$, $m \in \mathbb{N}$, $t_1, t_2, \cdots, t_m \in \mathbb{R}$, $\varepsilon > 0$ 
\begin{equation*}
W(f, t_1, \cdots, t_m, \varepsilon) = \left\{ g \in C(\mathbb{R}) \mid |g(t_i) - f(t_i)|<\varepsilon \ \forall i=\{1,... ,m\} \right\}.
\end{equation*}
I want to show the following:
a) Show that the uniform convergence topology is finer than the pointwise convergence topology
b) Show that the uniform convergence topology does not have a countable basis. 
c) Show that in the uniform convergence topology all sets of bounded and unbounded functions are closed and open sets at the same time.  
 A: a) Suppose that $O$ is open in the pointwise convergence topology.
Let $f \in O$. Then there is some $\varepsilon > 0$ and points $t_1, \ldots, t_n \in \mathbb{R}$ such that $W(f, t_1, \ldots t_n; \varepsilon) \subseteq O$.
Then $V(f, \varepsilon) \subseteq W(f, t_1, \ldots t_n; \varepsilon) \subseteq O$ as well ($g \in V(f,\varepsilon)$ implies that $|g(t_i) - f(t_i)| \le \sup_{t \in \mathbb{R}} |f(t) - g(t)| < \varepsilon$ for all $i=1, \ldots n$) and so $x$ is an interior point of $O$ in the uniform convergence topology as well.
As this holds for all $x \in O$. $O$ is open in the uniform convergence topology.
b) If $C(\mathbb{R})$ in the uniform convergence topology would have a countable base, so would all subspaces have. But for each subset $A$ of $\mathbb{Z}$ we can find a continuous function $f_A: \mathbb{R} \to \mathbb{R}$ such that for all $n \in \mathbb{Z}$: $f_A(n) = 1$ for $n \in A$ and $f_A(n) = 0$ for $n \not\in A$. Then $f_B \notin V(f_A, 1)$ for any distinct subsets $A,B$ of $\mathbb{Z}$, so $\{f_A: A \subseteq \mathbb{Z}\}$ is an uncountable discrete subspace (so cannot be second countable or even separable).
c) If $f$ is a bounded function, $V(f,1)$ only contains bounded functions too:
if for all $x \in \mathbb{R}$ we have $|f(x)| \le B$ for some bound $B$, then for any $g \in V(f,1)$ we can use the bound $B+1$. So the set of bounded functions is open (each point is an interior point).
And if $f$ is an unbounded function, so are all functions in $V(f,1)$. If some $g \in V(f,1)$ would have a bound $B$, then again we could bound $f$ by $B+1$ which we cannot. So also the unbounded functions are open. As these sets are each other's complement they are also closed. It shows that $C(X)$ in the uniform convergence topology is disconnected, as both sets are clearly non-empty $(\forall x: f(x) = x$ vs $\forall x:f(x) =1$). 
It's not so hard to see that the pointwise convergence topology is just the product topology on $\mathbb{R}^{\mathbb{R}}$ restricted to $C(\mathbb{R})$.
