I tried to solve it using the triangle inequality. I ended up with a $\delta$ that wasn't strictly positive, so the proof doesn't work. Here was the attempt:
Fix $\epsilon>0$. Suppose $1>-x>0$. Then $8>3-5x>3$ and so $8>|5x-3|$. Then $|5x^2-3x+\sin(x)| \leq |5x^2-3x|+|\sin(x)|$ by the triangle inequality and $|\sin(x)| \leq 1$ so $|5x^2-3x+\sin(x)| \leq |x||5x-3|+1<8|x|+1$.
In a perfect world at this point I would choose $\delta = \min \{ \frac{\epsilon-1}{8}, 1\}$ and conclude that $\epsilon=8\frac{\epsilon-1}{8}+1>8|x|+1 > |5x^2-3x+\sin(x)|$. However, $\delta>0$ won't hold when $\epsilon \leq 1$, so the proof doesn't work. Is there a way to prove it by continuing this line of reasoning, or should I prove it in some other way (if s0, how)?
Your upper bound for $|\sin(x)|$ is far too large. We know $\sin(x)\to 0$ as $x\to 0$ so for some $\delta_1>0$, we have $|x|<\delta_1$ implies $|\sin(x)|<\varepsilon$
So using your reasoning we can let $\delta_0 = \frac{\varepsilon}{8}$ and the $\delta_1$ as above and choose $\delta = \min\{\delta_0, \delta_1\}$ so that $|x|<\delta$ (and thus $|x|<\varepsilon/8$ and $|x|<\delta_1$) implies,
$$ |5x^2-3x+\sin(x)|\leq |x||5x-3|+|\sin(x)| < 8\delta + \varepsilon < \varepsilon + \varepsilon = 2\varepsilon $$
I usually don't care if there is an extra coefficient with the $\varepsilon$ but this is easily fixed by dividing things by 2