Scratchwork:
$$\forall x,y\in(0,1),\,x+y\gt x\sqrt{1-y^2}+y\sqrt{1-x^2}\implies\arcsin(x+y)\gt\arcsin(x)+\arcsin(y)$$
When $-1\le x+y\le1$. Wlog let $y\gt x$. Then:
$$\arcsin(|\sin(1/x)-\sin(1/y)|)=\arcsin(\sin(1/x)-\sin(1/y))\gt\frac{1}{x}-\frac{1}{y}=\frac{y-x}{xy}\gt\frac{y-x}{y^2}$$
If we let this be greater than $\arcsin(\varepsilon)>0$, we have that $\frac{y-x}{\arcsin(\varepsilon)}\gt y^2$. This gives $y=\sqrt{\frac{y-x}{\arcsin\varepsilon}}\lt\sqrt{\frac{\delta}{\arcsin\varepsilon}}$ when $y-x\lt\delta$. To disprove uniform continuity, it is sufficient to show that for some $\varepsilon\gt0$ we have $\forall\delta\gt0,\,\exists y,x:|y-x|\lt\delta$ but $|f(x)-f(y)|\gt\varepsilon$. If this can be shown for $\delta\in(0,a)$, then it holds for $\delta\in(0,b)$ for all $b\gt a$ as well, so considering the narrowed range of $\delta\in(0,\arcsin(\varepsilon))$ is sufficient.
Proof:
Let $1\gt\varepsilon\gt0$ and then let $\delta\in(0,\arcsin(\varepsilon))\neq\emptyset$. Let $y\in(0,\frac{1}{2}\sqrt{\frac{\delta}{\arcsin\varepsilon}})\subset(0,1)$ and let $x\in(y-\delta,y-\frac{\delta}{4})\neq\emptyset$, such that $-1\le\sin(1/x)-\sin(1/y)\le1$. Then $1\gt\delta\gt y-x\gt\frac{\delta}{4}\gt0$ and $\arcsin$ is well defined for $y-x$, and importantly $|y-x|\lt\delta$.
$$y\lt\frac{1}{2}\sqrt{\frac{\delta}{\arcsin\varepsilon}}\implies y^2\lt\frac{1}{4}\frac{\delta}{\arcsin\varepsilon}\implies\arcsin\varepsilon\lt\frac{\delta}{4y^2}\lt\frac{y-x}{y^2}\lt\frac{y-x}{xy}=\frac{1}{x}-\frac{1}{y}$$
By the increasing property of sine in $(0,1)$ and what was said during the scratchwork:
$$\varepsilon\lt\sin\left(\frac{1}{x}-\frac{1}{y}\right)\lt\sin\left(\frac{1}{x}\right)-\sin\left(\frac{1}{y}\right)$$
Despite $|y-x|\lt\delta$ for all $\delta\gt0$. As $\varepsilon$ was arbitrary, this concludes the proof that $\sin(1/x)$ is not uniformly continuous in $(0,1)$.
Is this correct? Is there a much neater approach?
It's a bit laborious $-$ you don't need all that scratchwork and all those arcsins. I would say:
Let $\epsilon=1$. Then every pair of the form $(x_n,y_n)=\left(\frac{1}{(2n+\frac12)\pi},\frac{1}{(2n-\frac12)\pi}\right)$ satisfies $|f(x_n)-f(y_n)|=2>\epsilon$. But for any $\delta>0$, we can choose $n$ large enough so that $|x_n-y_n|<\delta$.