LUXEMBOURG -- Switzerland's Timea Bacsinszky defeated fifth-seeded Belgian Yanina Wickmayer, 3-6, 6-2, 7-5 on Saturday to make the finals of the Luxembourg Open.
Germany's Sabine Lisicki is the other finalist after coming back to defeat Israel's Shahar Peer 6-3, 4-6, 7-6 (5).
The unseeded Bacsinszky ended Wickmayer's eight-match winning streak.
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Bacsinszky had lost to Wickmayer last week in Linz, Austria, where Wickmayer won her second career WTA Tour title.
This time, the Swiss player rallied from a set down and blew a 4-1 lead in the third set but held on. Bacsinszky will play for her first title.
The sixth-seeded Lisicki fought back from 4-2 down in the third set then 5-3 down in the tiebreaker.
Lisicki, who won her first career title at Charleston in April, beat Bacsinszky in their only previous match in this year's Fed Cup.
