Is this proof correct?
Let $\phi(x) > 0$ be a continuous function on $\mathbb{R}$ with $\lim_{x\rightarrow\pm\infty} \phi(x) = 0$ and let $f_n$ be an equicontinuous sequence of functions on $\mathbb{R}$ satisfying $\lvert f_n(x)\rvert \leq \phi(x)$ for all $x\in \mathbb{R}$ and $n\in\mathbb{N}$. Prove that $f_n$ has a uniformly convergent subsequence.
Since $\lim_{x\rightarrow\pm\infty} \phi(x) = 0$ there exists an $R \in \mathbb{R}$ such that if $\lvert x \rvert > R$ then $\phi(x) < 1$. Then $\phi(x)$ attains a maximum on $[-R, R]$, say M, as $\phi$ is continuous there. So $\max_{x\in\mathbb{R}} \phi(x) \leq \max(M, 1) = K$.
Since $\lvert f_n(x) \rvert \leq \phi(x) \leq K$ for all $x\in \mathbb{R}$ and $n\in\mathbb{N}$, we have that $f_n$ is uniformly bounded.
So $f_n$ is uniformly bounded and equicontinuous and so has a uniformly convergent subsequence $g_n$ on $[-\alpha, \alpha]$ for all $\alpha \in \mathbb{R}$.
Now, let $\epsilon > 0$ and choose $R \in \mathbb{R}$ such that $\phi(x) < \frac{\epsilon}{2}$ when $\lvert x\rvert > R$, and choose an $N$ large enough that when $n,m > N$ $\max_{\lvert x\rvert \leq R}\lvert g_n(x) - g_m(x)\rvert < \epsilon$ as $g_n$ is uniformly convergent on $[-R,R]$ and hence is uniformly Cauchy there. Then when $n, m > N$ we have that $$\max_{x\in\mathbb{R}}\lvert g_n(x) - g_m(x)\rvert = \max(\max_{\lvert x\rvert \leq R}\lvert g_n(x) - g_m(x)\rvert, \max_{\lvert x\rvert>R}\lvert g_n(x) -g_m(x)\rvert) < \max(\epsilon, 2\frac{\epsilon}{2}) = \epsilon$$ and so $g_m$ is actually uniformly Cauchy on all of $\mathbb{R}$. Hence $f_n$ has a uniformly convergent subsequence.