We define the following set $A\subset C(\mathbb{R})$ of continuous functions:
$f\in A$ if there exist a sequence $(s_{i})_{i\in\mathbb{Z}}$ of real numbers such that
(a) $\dots s_{-4}<s_{-2}<0\leq s_{0}<s_{2}<s_{4}<\dots,$
(b) $\lvert s_{2i+2}-s_{2i}-1 \rvert \leq \dfrac{1}{2}, $
(c) $s_{2i+1}=(s_{2i}+s_{2i+2})/2, \: i\in\mathbb{Z};$
and a sequence of natural numbers $(n_{i})_{i \in\mathbb{Z}}$ such that
(d) $f(s_{2i})=n_{i}$,
(e) $f(s_{2i+1})=0$,
(f) $f''|_{]s_{i},s_{i+1}[}\equiv 0$ for all $i\in\mathbb{Z}$
We write $f_{(s_{2k},n_{k})_{k}}$ to denote the continuous function $f$ associated with sequences $(s_{2k})_{k\in\mathbb{Z}}$ and $(n_{k})_{k\in\mathbb{Z}}$ given above.
$A$ is a closed subset of $C(\mathbb{R})$, where $C(\mathbb{R})$ is endowed with the compact open topology (that is, the topology of uniform convergence on compact subsets of $\mathbb{R}$), therefore a complete separable metric space.
Indeed, if, if $(f_{j})_{j}$ is a sequence of functions in $A$ that converges to certain $f\in C(\mathbb{R})$, then each $f_{j}$ has associated sequences $\left(s_{2k}(j)\right)_{k\in \mathbb{Z}}$ and $\left(n_{k}(j)\right)_{k\in \mathbb{Z}}$ satisfying conditions $(a)-(f)$ above.
I don't understand why : From the convergence with respect to the compact-open topology we deduce that there exist the limits $\lim_{j}s_{2k}(j)$ and $\lim_{j}n_{k}(j)$ for each $k\in \mathbb{Z}$ ? and also $f\in A$ ?
my attempt:
I know we have to use the definition of topology of uniform convergence on compact subsets of $\mathbb{R}$, $\forall K$ compact $$\lim_{j\to \infty}\sup_{x\in \mathbb{R}}\lVert f_{(s_{2k}^{j},n_{k}^{j})}^{j}(x) -f(x)\rVert=0 \qquad *$$
if we replace $x$ by $s_{2k}^{j}$ to use (d) we can not conclude because we will find $s_{2k}^{j}$ in both terms of $\quad$*
From the inequalities for the $s_i$ it follows that $s_{2i}\le \frac32(i+1)$ for $i\ge0$ and $s_{2i}\ge-\frac32i$ for $i<0$. This means that for every $i$ the sequence $\langle s_i(j)\rangle$ has a convergent subsequence.
In addition, working on a compact interval $K=[-\frac32i,\frac23(i+1)]$ and using that the limit function $f$ is bounded on $K$ you will find that $\langle n_{2i}(j)\rangle$ takes on finitely many values.
Now using these two facts and recursion you can find a subsequence $\langle f_{j_k}\rangle$ of your given sequence such that all sequences $\langle s_i(j_k)\rangle$ and $\langle n_{2i}(j_k)\rangle$ converge. Take an infinite set $A_0$ such that $\langle s_0(j)\rangle_{j\in A_0}$ converges, then an infinite subset $B_0$ of $A_0$ such that $\langle n_0(j)\rangle_{j\in B_0}$ is constant, then an infinite subset $A_1$ of $B_0$ such that $\langle s_{-2}(j)\rangle_{j\in A_1}$ converges, then $\ldots$. In the end take an infinite set $X$ such that $X\setminus A_i$ is finite for all~$i$. Then $\langle f_j\rangle_{j\in X}$ is the desired sequence.