Proof that the topology of uniform convergence on $C(\mathbb R, X)$ is finer than the topology of pointwise convergence. The proof is from R. Engelking, General Topology and it relies on the following fact
$$
  f \textrm{ is continuous} \Leftrightarrow f(\overline{A}) \subseteq \overline{f(A)}. \qquad (1)
$$
Proposition: For every topological space $X$ the topology of uniform convergence on $C(\mathbb R, X)$ is finer than the topology of pointwise convergence.
Proof: The equivalence of (1) above shows that it suffices to prove that if $f \in C(\mathbb R, X)$ is in the closure of a set $A \subseteq C(\mathbb R, X)$ with respect to the topology of pointwise convergence, then $f$ is in the closure of $A$ with respect to the topology of pointwise convergence. Let $U = C(\mathbb R, X) \cap \bigcap_{i=1}^k p_{x_i}^{-1}(U_i)$ be a neighboorhood of $f$ in the topology of pointwise convergence; since the sets $U_i$ are open in $R$, there exists an $\epsilon > 0$ such that $(f(x_i) - \epsilon, f(x_i) + \epsilon) \subseteq U_i$ for $i = 1,2,\ldots,k$. As $f = \lim f_j$, where $f_j \in A$ for $j = 1,2,\ldots$ there exists a $j$ such that $|f(x) - f_j(x)| < \epsilon$ for every $x \in X$, in particular, $f_j(x_i) \in U_i$ for $i = 1,2,\ldots, k,$ and this shows that $U\cap A \ne \emptyset$. q.e.d.
I don't understand the proof strategy, why "The equivalence of (1) above shows that it suffices to prove that if $f \in C(\mathbb R, X)$ is in the closure of a set $A \subseteq C(\mathbb R, X)$ with respect to the topology of pointwise convergence", I don't see the connection, because it is a characterisation of continuous functions, and has so nothing to do with the topology on the set $C(\mathbb R, X)$ (because here continouity refers to the topologies on $\mathbb R$ and $X$), can someone explain this to me?
 A: It's a typo, it should read

The equivalence of (1) above shows that it suffices to prove that if $f \in C(\mathbb R, X)$ is in the closure of a set $A \subseteq C(\mathbb R, X)$ with respect to the topology of uniform convergence, then $f$ is in the closure of $A$ with respect to the topology of pointwise convergence.

The connection to $(1)$ is that $\mathcal{T_1} \supset \mathcal{T}_2 \iff \operatorname{id} \colon (X,\, \mathcal{T}_1) \to (X,\, \mathcal{T}_2)$ is continuous; and the inclusion of the closure with respect to uniform convergence in the closure with respect to pointwise convergence is, in light of $(1)$, just what's needed to show the continuity of $\operatorname{id}\colon \bigl(C(\mathbb R, X),\, \mathcal{T}_{uni}\bigr) \to \bigl(C(\mathbb R, X),\, \mathcal{T}_{pw}\bigr)$.
Note however, that here it is much simpler to directly prove that every neighbourhood of $f$ in the topology of pointwise convergence contains a neighbourhood of $f$ in the topology of uniform convergence,
$$\{g \in C(\mathbb R, X)\colon \lvert g(x_i) - f(x_i)\rvert < \varepsilon_i\} \supset \{h\in C(\mathbb R, X)\colon \sup_x \lvert h(x) - f(x)\rvert < \min_i \varepsilon_i\}.$$
