How to skip app select when clicking on a link

I’ve successfully set up deep linking for my application, but in some cases I would like to open link which is exactly the same as the one which opens my app.

For example:
example.com/course// # This opens course screen in application

How can I add a button in my app to open this link in browser if I selected to "Always open in ", when opened this link previous time? Now if I can click on this button nothing happens:

Linking.openURL(`https://example.com/course/${this.props.courseSlug}`)}

and

await WebBrowser.openBrowserAsync(`https://example.com/course/${this.props.courseSlug}`)}

Not working