fix: use os.startfile on Windows instead of subprocess start

[MAXDVVV]

Problem

click.launch("https://example.com") fails on Windows with [WinError 2] The system cannot find the file specified since version 8.1.8.

This is because start is a Windows shell built-in command, not an executable file. The migration from os.system() to subprocess.call() in #1477 broke this — subprocess.call(["start", ...]) requires shell=True to work, but the maintainers correctly don't want to use shell=True.

Fix

Replace subprocess.call(["start", ...]) with os.startfile(), which is the native Python API for opening files and URLs on Windows. It calls the Windows ShellExecuteW API directly without needing a shell.

• Non-locate case: Uses os.startfile(url) directly
• Locate case: Keeps subprocess.call(["explorer", "/select,..."]) (explorer.exe is a real executable)
• Fallback: If os.startfile fails with OSError, falls back to webbrowser.open() for HTTP(S) URLs (same pattern as the Linux/xdg-open fallback)

Test

Added test_open_url_windows_uses_startfile — monkeypatches os.startfile and verifies it's called with the URL.

Fixes #2868