Repository: vscode
Issue: VS Code web: quickpick disappear when was opened by a walkthrough button
Labels: feature-request,getting-started

There is similar/duplicating issue which has been closed without any resolution https://github.com/microsoft/vscode/issues/204522
I've just tested it on the GitPod with VS Code web 1.90.0 and can confirm the bug is still there.

What's the problem?

I added a button to the walkthrough and implemented reading of two values using `vscode.window.showInputBox` and `vscode.window.showQuickPick`.
The first method returns the actual entered value. Then after closing the input box, a quickpick appears and then immediately closed because the focus is returned to the walkthrough. For the quickpick this action is similar to pressing Esc or clicking the mouse somewhere outside, so it returns `undefined`.

There is a simple code demonstrating the bug

```javascript
        const port = await vscode.window.showInputBox({
            value: '8080',
            title: 'Exposed Port'
        });

        const exposure = await vscode.window.showQuickPick(['public', 'internal', 'none'], {
            title: 'Describe how the port should be exposed on the network'
        });

        await vscode.window.showInformationMessage(`Exposed port ${port}:${exposure}`);
```

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- VS Code Version: 1.90.0
- OS Version: web

Steps to Reproduce:

Here I want to pay attention that the bug is not reproducible permanently. It may happen that I cannot reproduce it at all. In such cases to reproduce the bug I have to play a bit with views (open and close) and then return to the walkthroug.

1. Go to GitPod and create a workspace using https://github.com/vitaliy-guliy/vscode-test-extension/tree/test-vscode-quickpick
2. Once the extension vsix file is created, install it.
3. Open `Test` walkthrough and click `Show` button. Then follow the popups.

A video demonstrating the problem

https://github.com/microsoft/vscode/assets/1655894/051e8814-3ca5-43a2-b4cd-92fb851bd2aa

Sure it is possible to add a small delay after reading the first value, but this will only hide the bug, not fix it
```javascript
    await new Promise<void>(resolve => setTimeout(resolve, 500));
```