Even without those simplifying assumptions, a bound can be obtained by combining a couple of usual tools:
- The variance of the difference of two correlated variables. It allows us to turn a two variables problem into an univariate problem.
- Chebyshev's inequality. It puts a bound on the probability of exceeding a given value.
In some detail:
$$\sigma^2_{X-Y}=\sigma^2_X+\sigma^2_Y-2·cov(X,Y)$$
$$cov(X,Y)=\sigma_X·\sigma_Y·\rho_{XY}$$
$$\sigma^2_{X-Y}=\sigma^2_X+\sigma^2_Y-2·\sigma_X·\sigma_Y·\rho_{X,Y}$$
According to Chebyshev's inequality, for any random variable $Z$:
$$ \Pr(|Z-\mu|\geq k\sigma) \leq \frac{1}{k^2}$$
Then (and using that $\mu_{X-Y}=\mu_X-\mu_Y)$:
$$ \Pr(|X-Y-\mu_X+\mu_Y|\geq k·\sqrt{\sigma^2_X+\sigma^2_Y-2·\sigma_X·\sigma_Y·\rho_{X,Y}}) \leq \frac{1}{k^2}$$
We can use the proposed simplifying assumptions to get a simpler expression. When:
$$\rho_{X,Y}= {covar(X,Y)} / {\sigma_X \sigma_Y} = 1 - \epsilon $$ $$\mu_x = \mu_y = 0$$ $$\sigma_X^2 = \sigma_Y^2 = \sigma^2$$
Then:
$$\sigma^2_X+\sigma^2_Y-2·\sigma_X·\sigma_Y·\rho_{X,Y} = 2·\sigma^2·(1-(1-\epsilon)) = 2\sigma\epsilon$$$$\sigma^2_X+\sigma^2_Y-2·\sigma_X·\sigma_Y·\rho_{X,Y} = 2·\sigma^2·(1-(1-\epsilon)) = 2\sigma^2\epsilon$$
And therefore:
$$\Pr(|X-Y|\geq k·\sigma\sqrt{2\epsilon}) \leq \frac{1}{k^2}$$
Interestingly, this result holds even if $\epsilon$ is not small, and if the condition for correlation changes from $=1-\epsilon$ to $\geq 1-\epsilon$, the result doesn't change (because it's already an inequality).