Subsequence Convergence If $\{f_{n}\}_{n\geq 1}$ is a sequence of continuous functions on $\mathbb R$, with $|f_{n}(x)|\leq 1$ for all $x\in \mathbb{R}$, and all $n\geq 1$. Does there exist a subsequence which converges uniformly (or pointwise) to some continuous function $f$?

As I know the Arzelà–Ascoli theorem works for closed intervals $[a,b]$, I don't know if there is something in case of $\mathbb{R}$? 

EDIT: If this assumption help we can consider it: the sequence $\{f'_{n}\}$ is uniformly bounded on $\mathbb R$.
Take $f_n(x) = x^n 1_{(0,1)}(x)+1_{[1,\infty)}(x)$. Let $x_n = \frac{1}{\sqrt[n]{2}}$, and note that $x_n \to 1$. Also note that $f_n(x_n) = \frac{1}{2}$, $f_n(1) = 1$, and for a fixed $x$, $f_n(x)$ is non-increasing, so for $m\geq n$, $f_m(x_n) \leq \frac{1}{2}$.

Consequently no subsequence can converge to a continuous function. To get a contradiction, suppose $f_n(x) \to f(x)$, where $f$ is continuous. We must have $f(1) = 1$, so choose $\delta&gt;0$ such that if $|x-1|&lt;\delta$, then $|f(x)-1|&lt;\frac{1}{4}$. Now choose $n$ such that $|x_n-1| &lt; \delta$. Then if $m \geq n$ we have $f_m(x_n) \leq \frac{1}{2}$, hence $f(x_n) \leq \frac{1}{2}$, which is a contradiction.

Just noticed that there was an update to the question.

Answer to modified question:

If the derivatives are uniformly bounded, then the functions are uniformly Lipschitz, hence equicontinous. Since the functions themselves are uniformly bounded, we can apply Arzela-Ascoli on nested closed intervals to get the desired result:

Let $I_1 = [-1,1]$, and apply Arzela-Ascoli to get a subsequence (ie, an infinite subset) $N_1 \subset \mathbb{N}$ along which $\{f_n\}_{n \in N_1}$ converges uniformly to $f:I_1 \to \mathbb{R}$. (Since the convergence is uniform, $f$ is continuous.)

Now suppose we have a subsequence $N_k \subset \mathbb{N}$, an interval $I_k=[-k,k]$, and a continuous limit function $f:I_k \to \mathbb{R}$. Consider the functions $\{f_n\}_{n \in N_k}$ on the interval $I_{k+1} = [-(k+1),k+1]$. Again, by Arzela-Ascoli, there is a subsequence $\mathbb{N}_{k+1} \subset \mathbb{N}_k$ along which the functions converge uniformly to a continuous function $\phi: I_{k+1} \to \mathbb{R}$. Since the functions converge to $f$ on the subsequence $\mathbb{N}_k$, it follows that $\phi(x) = f(x)$, $\forall x \in I_k$. Hence we may abuse notation slightly by dropping the symbol $\phi$ and using $f$ to denote the new limit function, ie, $f:I_{k+1} \to \mathbb{R}$.

Continuing this way, we define a continuous function $f:\mathbb{R} \to \mathbb{R}$. If $x \in \mathbb{R}$, then for sufficiently large $k$, we have $x \in I_k$, and $\{f_n(x)\}_{n \in N_k}$ converges to $f(x)$.

The convergence is uniform on any bounded set (since a bounded set will be contained in $I_k$ for some $k$). The convergence need not be uniform on all of $\mathbb{R}$, however. For example, take $f_n(x) = 1_{(-2 n \pi, 2 n \pi)}(x) + \cos(x) 1_{(-\infty,-2 n \pi] \cup [2 n \pi, \infty)}(x)$. Both $f_n$ and $f_n'$ are uniformly bounded. Clearly, $f_n$ converges pointwise to $f(x) = 1$, but $\max_{x \in \mathbb{R}} |f_n(x)-f(x)| = 2$, $\forall n$, and so the convergence in not uniform.
No, it doesn't exist. Something very bad could happen in the infinity.