Let $\{f_n\}_{n\in\mathbb{N}}\subset C\left(\mathbb{R}^2;\mathbb{R}\right)$ be a sequence of continuously differentiable, radial functions verifying $$ \tag{*}\label{*} \int_{\mathbb{R}^2}\left[(f_n(\vec{x}))^2+\|\nabla f_n(\vec x)\|^2\right]d\vec x\le1\qquad\forall n\in\mathbb{N}. $$ Then show that there exists $\{f_{n_k}\}_{k\in\mathbb{N}}\subset\{f_n\}_{n\in\mathbb{N}}$ converging uniformly on every compact subset of $\mathbb{R}^2\backslash\{0\}$.
It is suggested to use the theorem of Arzela-Ascoli (which too at first sight seems to be the most reasonable approach). I already showed that it is enough to show that for all $r,R\in\mathbb{R}_{>0}$ with $r<R$ we can find a subsequence $\{f_{n_k}\}_{k\in\mathbb{N}}$ (depending on $r$ and $R$) uniformly converging over $K_{r,R}:=\overline{B(0;R)}\backslash B(0;r)$. So we need to verify that $\{f_{n}\}_{n\in\mathbb{N}}$ is uniformly bounded and equicontinuous on $K_{r,R}$, which I have done as written further down. My solution isn't very elegant and rather lengthy; I have a strong feeling that there is a more straightforward solution. If you have the time it would be great if you could verify that my solution is correct, but my main question is:
How to solve it more elegantly?
Let $n\in\mathbb{N}$, so as $f_n$ is radial there exists $r_n\in C^1(\mathbb{R}_{≥0};\mathbb{R})$ with $f_n(\vec x)=r_n\left(\|\vec x\|\right)$ for all $\vec x\in\mathbb{R}^2$ (we also notice that $r'_n(0)=0$ but that's not so important). Firstly we see that in radial coordinates $$ \int_{K_{r,R}}\left[(f_n(\vec{x}))^2+\|\nabla f_n(\vec x)\|^2\right]d\vec x=2\pi\int_r^R\rho\left[(r_n(\rho))^2+(r'_n(\rho))^2\right]d\rho\\ ≥2\pi r\int_r^R\left[(r_n(\rho))^2+(r'_n(\rho))^2\right]d\rho $$ and thus $$ \tag{**}\label{**} \int_r^R\left[(r_n(\rho))^2+(r'_n(\rho))^2\right]d\rho\stackrel{\eqref{*}}≤\frac{1}{2\pi r}. $$ Now let $M=1+\frac{2\max\{1,R-r\}}{\sqrt{2\pi r(R-r)}}$ and suppose $|r_n(r)|>M$. By replacing $r_n$ through $-r_n$ (which doesn't influence $\eqref{**}$) we can assume $r_n(r)>0$. Suppose that $r_n(x)≥\frac{M}{2}$ for all $x\in[r;R]$, then $$ (R-r)\frac{M^2}{4}≤\int_r^R(r_n(\rho))^2d\rho≤\int_r^R\left[(r_n(\rho))^2+(r'_n(\rho))^2\right]d\rho\stackrel{\eqref{**}}≤\frac{1}{2\pi r}\implies\\ M≤\frac{2}{\sqrt{2\pi r(R-r)}} $$ which is a contradiction. Thus, by the intermediate value theorem there exists $x\in]r;R]$ with $r_n(x)=\frac{M}{2}$. But then $$ \frac{1}{2\pi r}\stackrel{\eqref{**}}≥\int_r^R\left[(r_n(\rho))^2+(r'_n(\rho))^2\right]d\rho≥\int_r^x(r'_n(\rho))^2d\rho\stackrel{CS}≥\frac{1}{x-r}\left(\int_r^x r_n'(\rho)d\rho\right)^2=\frac{M^2}{4(x-r)}\implies\\ M≤\frac{2\sqrt{x-r}}{\sqrt{2\pi r}}≤\frac{2\sqrt{R-r}}{\sqrt{2\pi r}}=\frac{2(R-r)}{\sqrt{2\pi r(R-r)}} $$ which again is a contradiction (CS stands for Cauchy-Schwarz). Thus $|r_n(r)|≤M$. Finally we see that for $x\in[r;R]$ we have $$ (r_n(x))^2-(r_n(r))^2=\int_r^x 2r_n(\rho)r_n'(\rho)d\rho≤\int_r^R\left[(r_n(\rho))^2+(r'_n(\rho))^2\right]d\rho\stackrel{\eqref{**}}≤\frac{1}{2\pi r}\implies\\ |r_n(x)|≤\sqrt{\frac{1}{2\pi r}+(r_n(r))^2}≤\sqrt{\frac{1}{2\pi r}+M^2}, $$ therefore $\{r_n\}_{n\in\mathbb{N}}$ is uniformly bounded on $[r;R]$ and thus also $\{f_n\}_{n\in\mathbb{N}}$ on $K_{r,R}$.
For the equicontinuity over $K_{r,R}$ it again suffices to prove that $\{r_n\}_{n\in\mathbb{N}}$ is equicontinuous over $[r,R]$. In order to do so, let $\varepsilon>0$ and $\delta=2\pi r\varepsilon^2$, then for $x,y\in[r;R]$ with $|x-y|<\delta$ (and WLOG $y≥x$) we see that $$ |r_n(y)-r_n(x)|=\left|\int_x^y r_n'(\rho)d\rho\right|≤\int_x^y |r_n'(\rho)|d\rho\stackrel{CS}≤\left((y-x)\int_x^y(r_n'(\rho))^2d\rho\right)^{\frac{1}{2}}≤\left(\delta\int_r^R\left[(r_n(\rho))^2+(r'_n(\rho))^2\right]d\rho\right)^{\frac{1}{2}}\stackrel{\eqref{**}}≤\left(\frac{\delta}{2\pi r}\right)^{\frac{1}{2}}=\varepsilon $$ which proves the equicontinuity.
If you prove equicontinuity first (I think your proof of it is the most direct one), then equiboundedness follows from it.
Fix $0<r<R$. For every $r_n$ there is a point $x_n\in(r,R)$ such that $r_n(x_n)^2\leq \frac{1}{2\pi^2r(R^2-r^2)}$, otherwise we would have $$\int_r^R {r_n}^2+{r'_n}^2\geq \int_r^R {r_n}^2>\pi(R^2-r^2)\frac{1}{2\pi^2r(R^2-r^2)}=\frac{1}{2\pi r}$$ which contradicts $(**)$.
Now use equicontinuity, choosing $\varepsilon$ such that $\delta(\varepsilon)=R-r$, which from your proof can be $\varepsilon=\frac{1}{\sqrt{2\pi(R-r)}}$. Then for any $y\in (r,R)$ we have $$|r_n(y)|\leq |r_n(y)-r_n(x_n)|+|r_n(x_n)|\leq \varepsilon+|r_n(x_n)|\leq \frac{1}{\sqrt{2\pi(R-r)}}+\frac{1}{\pi\sqrt{2r(R^2-r^2)}}$$ since $|y-x_n|<R-r=\delta(\varepsilon)$.
This is essentially the idea you used, but from what I see you were kind of reproving equicontinuity at each step, which made the proof lengthy.