Looking at the asymptotic expansion of the Bessel function of the first kind $J_0(x)$, the following bound seems an obvious conjecture
$$ \left | J_0(x) \right | \le \min \left (1, \sqrt{\frac{2}{\pi x}} \right) $$
I would also be interested in any bound of the form $J_0(x) \le A/ \sqrt{x +B}$ for some constants $A>0,\, B\ge0$. The bound looks tight when plotted on short segments and seems an obvious result. However I was a bit surprised when coming across this article.
Notes that the bounds are meant to hold for all real $x$, though $J_0(x)$ is even so one can restrict oneself to $x\ge0$.
This bound is known, and one citation given for it is G.N. Watson's book A Treatise On The Theory Of Bessel Functions, pp 205-8. Even accounting for the book's quite unfriendly writing style, it's not clear to me that the bound actually occurs therein.... Here is a complete proof.
Modulo a computation on the interval $[0,2]$, it suffices to show that $\frac{\pi x}2 J_0(x)^2 \le 1$ for $x\ge2$. At the top of page 206, equation (1), we have $$ J_0(x) = \sqrt{\frac2{\pi x}} \big( \cos(x-\tfrac\pi4) P(x,0) - \sin(x-\tfrac\pi4) Q(x,0) \big), $$ where $P$ and $Q$ are certain functions. Then on page 208, we take $p=2$ in equation (1) to get $$ 1 - \frac{1^23^2}{2!(8x)^2} \le P(x,0) \le 1 - \frac{1^23^2}{2!(8x)^2} + \frac{1^23^25^27^2}{4!(8x)^4}, $$ and we take $p=0$ in equation (2) to get $$ -\frac1{8x} \le Q(x,0) \le 0. $$ Therefore by Cauchy's inequality, \begin{align*} \frac{\pi x}2J_0(x)^2 &\le \big( \cos(x-\tfrac\pi4)^2 + \sin(x-\tfrac\pi4)^2 \big) \big( P(x,0)^2 + Q(x,0)^2 \big) \\ &\le 1 \cdot \bigg( \bigg( 1 - \frac{1^23^2}{2!(8x)^2} + \frac{1^23^25^27^2}{4!(8x)^4} \bigg)^2 + \frac1{(8x)^2} \bigg) \\ &\le \bigg( \bigg( 1 - \frac{1^23^2}{2!(8x)^2} + \frac{1^23^25^27^2}{4!(8x)^216^2} \bigg)^2 + \frac1{(8x)^2} \bigg) \\ &= 1-\frac{4517}{65536 x^2}+\frac{30702681}{17179869184 x^4}, \end{align*} where the final inequality used $x\ge2$. It's now easy to verify that this last bound is less than $1$ for $x\ge2$.
As for the interval $[0,2]$: for $x$ in this interval, the series $$ J_0(x) = \sum_{m=0}^\infty \frac{(-1)^m}{m!^2} \bigg(\frac x2 \bigg)^2 $$ satisfies the hypotheses of the alternating series test, and thus the series's truncations alternately form upper and lower bounds for $J_0(x)$ on this interval. The first two truncations give $1 \ge J_0(x) \ge 1-x^2/4 \ge 0$. The next truncation gives $$ J_0(x) \le 1-\frac{x^2}4 + \frac{x^4}{64} < \sqrt{\frac2{\pi x}}, $$ where we need to verify the last inequality to finish the proof. Taking square roots of both sides and subtracting, it suffices to show that $$ \sqrt[4]{\frac2{\pi x}} - \bigg( 1-\frac{x^2}8 \bigg) > 0 $$ on $[0,2]$. Standard calculus shows that the minimum of this function occurs at $x=(2/\pi)^{1/9}$ and has value $9/(2^{25/9} \pi ^{2/9})-1$; verifying that this minimum value is positive boils down to confirming that $\pi < 3^9/2^{25/2}$, which in turn (since $\pi<22/7$) follows from the inequality $3^{18}7^2 > 2^{27} 11^2$.