Courtney Vandersloot bids Sky farewell, looks ahead to ‘new beginning’

Courtney Vandersloot announced in a social media post Tuesday that she would not be returning to the Chicago Sky, setting the stage for her free move to another WNBA team.

Free agents can officially sign with teams starting Wednesday.

Vandersloot, the No. 3 overall Gonzaga in 2011, is one of the best point guards in WNBA history, having led the league in assists per game six times in her 12 seasons in Chicago. She also helped guide Sky to a WNBA championship in 2021.

“To the Sky organization who pulled in the little guard from a midfielder and believed in me from the jump I couldn’t have asked for anything better,” Vandersloot wrote on Instagram. “Thank you from the bottom of my heart. I made my dreams come true because of you.

“Even though I never planned this day, I’ve decided it’s time to make a fresh start. I will forever be grateful for the memories I made during my time here. As I look forward to a new chapter, with a new team, in a new city, knowing that Chicago, its fans and the Sky organization will always hold a special place in my heart. It’s goodbye for now, but thank you forever.”

Since the end of last season, when Sky lost to the Connecticut Sun in the semifinals, there has been speculation that Vandersloot, an unrestricted free agent, could join from Chicago this year. Her decision could be partly tied to another free agent, forward Breanna Stewart, who is deciding between staying with the Seattle Storm and joining the New York Liberty.

Vandersloot and Stewart could choose to play together.

Vandersloot, who turns 34 on February 8, has averaged 10.2 points and 6.6 assists over her WNBA career. Her wife, guard Allie Quigley, is also an unrestricted free agent, having spent the last 10 seasons in Chicago.

Another player from Sky’s championship squad, Candace Parker, took to social media on Saturday to announce that she would be joining the Las Vegas Aces. Courtney Vandersloot bids Sky farewell, looks ahead to ‘new beginning’

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