= Solution
<sulfide> first opens one <epoxide> by backside attack, giving a carbon-bound <thiolate> and an <alkoxide>. Using the same numbering, take this first attack at C2: the C2–S bond is down and C3–O remains up.
The attached <sulfur> still bears a nucleophilic <lone pair> and negative charge. It attacks C5 of the second <epoxide> \b[intramolecularly], breaking C5–O and closing a bridge between C2 and C5. Work-up protonates the <oxygen> anions at C3 and C6. Thus \b[B is 7-thiabicyclo$[2.2.1]$heptane-2,5-diol], with both OH groups on the face opposite the <sulfur> bridge.
Both C–S bonds are formed by inversion at their attacked <epoxide> <carbons>. Because the original <epoxides> are cis, both new C–S bonds point to the same opposite face and one <sulfur> <chemical atom> can connect them. The 1,4 bridge gives two five-membered sulfur-containing circuits, unlike the more strained closure between nearby 1,3 positions.
The <sulfur> therefore replaces two separate external <nucleophiles>: \b[two <epoxides> are opened but only one <sulfur> <chemical atom> is incorporated]. Adding that <sulfur> and two work-up <protons> gives $\mathrm{C_6H_{10}O_2S}$, consistent with B.
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