This is a fun one, and a common question in this course. First, as you say, H and Br are lost by anti elimination, since they helpfully tell you base and E2 are involved. That means the H and Br are lost from opposite faces of the future double bond, leaving us with a trans intermediate that has a mirror plane (the plane of the molecule)
Now we need to add Br+ to make a bromonium intermediate. Notice that both carbons end up as (R) or else both end up as (S), so two enantiomers are possible here.
But, whichever carbon is attacked by Br- ends up with an inversion of symmetry, so you now have two carbons of opposite chirality. The result, as shown in your answer, is a molecule with a mirror plane. In other words, adding bromine to opposite sides of a trans double bond makes both sides come out as mirror images, once you turn the bromines around to line up. This compound is a meso compound. But ... what about the other isomer if the Br+ originally goes to the other side of the double bond? Well, just turn your Fisher structure upside down. :)