Alright, you say known variances. So it's an exercise on a point of theory, not a realistic problem.
And you actually assume the two sample sizes are equal.
Start by recalling something from the one-sample problem: $$ \bar{X} = \frac{X_1+\cdots+X_n}{n} \sim N\left(\mu_X,\frac{\sigma^2_X}{n} \right) $$ $$ \bar{Y} = \frac{Y_1+\cdots+Y_n}{n} \sim N\left(\mu_Y,\frac{\sigma^2_Y}{n} \right) $$ You don't explicitly state that the two samples are independent. If they are, they we have $$ \bar X - \bar Y \sim N\left(\mu_X-\mu_Y,\frac{\sigma^2_X+\sigma^2_Y}{n}\right) $$ (If we had unequal sample sizes $n$ and $m$, then the variance would be $\dfrac{\sigma^2_X}{n}+\dfrac{\sigma^2_Y}{m}$.)
So $$ \frac{((\bar X-\mu_X) - (\bar Y-\mu_Y))\sqrt{n}}{\sqrt{\sigma^2_X+\sigma^2_Y}} \sim N(0,1). $$ So the probability that $$ -A < \frac{(\bar X-\mu_X) - (\bar Y-\mu_Y)}{\sqrt{ \frac{\sigma^2_X+\sigma^2_Y}{n} }} <A \tag{1} $$ is the desired confidence when the number $A$ is suitably chosen. Now do a bit of algebra to rearrange the inequalities $(1)$: $$ \bar X - \bar Y - A\sqrt{\frac{\sigma^2_X+\sigma^2_Y}{n}} < \mu_X-\mu_Y < \bar X - \bar Y + A\sqrt{\frac{\sigma^2_X+\sigma^2_Y}{n}} $$ That's the confidence interval.