LUXEMBOURG -- Surprise U.S. Open semifinalist Yanina Wickmayer beat Kirsten Flipkens 7-6 (1), 4-6, 6-1 in an all-Belgian quarterfinal at the Luxembourg Open on Friday.
Wickmayer, ranked 20th in the world, is seeking her second tournament victory in a row after winning last week's title in Linz, Austria. She will play 70th-ranked Timea Bacsinszky of Switzerland, who beat Katarina Srebotnik of Slovenia 6-7 (3), 6-1, 6-4.
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Seeded fifth in the tournament, Wickmayer became the highest seed in the tournament after Israel's Shahar Peer upset fourth-seeded Daniela Hantuchova of Slovakia 6-2, 7-6 (4).
Fans had been anticipating an all-Belgian final involving second-seeded Kim Clijsters, but the U.S. Open champ lost her second-round match against Patty Schnyder of Switzerland on Thursday.
That match left Schnyder tired, and it showed against sixth-seeded Sabine Lisicki. The German had no trouble in ousting the 30-year-old veteran 6-4, 6-2, and will now face Peer in Saturday's semifinals.
