Question on uniform convergence of a subsequence Any help would be great.

Let $F,G$ : $\mathbb{R}\to (0,\infty ) $ be continuous. Let $f_n: \mathbb{R} \to \mathbb{R} $ be everywhere differentiable and satisfy $ |f_n(x)| \leq F(x) $ and $ |f'_n(x)| \leq G(x)$ for all $x \in R$ and all $n \in \mathbb{N}$. Show that there is a subsequence $(f_{n_k})$ which converges uniformly on the compacta of $\mathbb{R}$ to some continuous function.

 A: We know $\mathbb{Q}$ is a countable set, so let $q_1,q_2,q_3,...$ be a sequence that enumerates all elements of $\mathbb{Q}$.
We know for all $n \in \mathbb{N}$  $|f_n(q_1)| \leq F(q_1)$ So bolzano-weierstrass theorem says there is a convergent sub-sequence, say $f_{k1_n}(q1)$ with limit $Q_{1}$ 
In the same way, for all $n \in \mathbb{N} |f_{k_n}(q_2)| \leq F(q_2) $. So there exists a convergent sub-sequence of $f_{k1_n}(q_2)$, say $f_{k2_n}(q_2)$ with limit $Q_2$
If we continue this process, we can generate a row of sub-sequences $(k1)_n \supseteq (k2)_n \supseteq (k3)_n \supseteq ...$ 
Now we define a sub-sequence of $f_n$, namely $g_n$ such that $g_n = f_{(kn)_n}$
Take $C$ , a random compact of $\mathbb{R}$, we will show that $g_n$ converges uniformly on $C$ to finish the proof.
$C$ is bounded, let $M$ be the maximal value of $G$ in a convex boundary of $C$. So $|F(x)-F(y)|<M|x-y|$ for x,y in $C$
Take $\epsilon >0$.
$\bigcup_n B(q_n,\frac{\epsilon}{4M})$ is a open cover of $C$ (because it is an open cover of $\mathbb{R}$), so because C is compact there exists a finite $Q \subset \mathbb{Q}$ such that $C \subset \bigcup_{q \in Q} B(q,\frac{\epsilon}{4M})$ let $q_r$ be the number of largest index of all numbers in $Q$. 
$\forall m \in \mathbb{N}$,$m>r ,\forall q_i \in Q : (k_m)_m \in (k_i)_n $ , so $(g_n(g_i))_n$ converges to $Q_i$. So we can choose $l \in \mathbb{N}$ sufficiently large such that $\forall q_i \in Q, \forall n \in \mathbb{N}: n>l : |g_n(q_i)-Q_i| < \epsilon/2 $ 
For any $x,y > l$ and $c \in C$ take $q_i \in Q$ such that $|c-q|<\epsilon/4M$ we have $|g_x(c) - g_y(c)| < |g_x(c) - g_x(q_i)| + |g_x(Q_i) - Q_i| + |Q_i - g_y(q_i)| + |g_y(q_i) - g_y(c)|$ $ < \frac{\epsilon}{4M}M + \frac{\epsilon}{4} + \frac{\epsilon}{4} + \frac{\epsilon}{4M}M = \epsilon$ 
This proves that $g_n$ is a Cauchy sequence for the uniform measure on $C$. Because of it's completeness we have: $g_n$ converges uniformly on $C$
