Setting DataContext in XAML in WPF

This code will always fail.

As written, it says: "Look for a property named "Employee" on my DataContext property, and set it to the DataContext property". Clearly that isn't right.

To get your code to work, as is, change your window declaration to:

<Window x:Class="SampleApplication.MainWindow"
    xmlns="http://schemas.microsoft.com/winfx/2006/xaml/presentation"
    xmlns:x="http://schemas.microsoft.com/winfx/2006/xaml"
    xmlns:local="clr-namespace:SampleApplication"
    Title="MainWindow" Height="350" Width="525">
<Window.DataContext>
   <local:Employee/>
</Window.DataContext>

This declares a new XAML namespace (local) and sets the DataContext to an instance of the Employee class. This will cause your bindings to display the default data (from your constructor).

However, it is highly unlikely this is actually what you want. Instead, you should have a new class (call it MainViewModel) with an Employee property that you then bind to, like this:

public class MainViewModel
{
   public Employee MyEmployee { get; set; } //In reality this should utilize INotifyPropertyChanged!
}

Now your XAML becomes:

<Window x:Class="SampleApplication.MainWindow"
        xmlns="http://schemas.microsoft.com/winfx/2006/xaml/presentation"
        xmlns:x="http://schemas.microsoft.com/winfx/2006/xaml"
        xmlns:local="clr-namespace:SampleApplication"
        Title="MainWindow" Height="350" Width="525">
    <Window.DataContext>
       <local:MainViewModel/>
    </Window.DataContext>
    ...
    <TextBox Grid.Column="1" Grid.Row="0" Margin="3" Text="{Binding MyEmployee.EmpID}" />
    <TextBox Grid.Column="1" Grid.Row="1" Margin="3" Text="{Binding MyEmployee.EmpName}" />

Now you can add other properties (of other types, names), etc. For more information, see Implementing the Model-View-ViewModel Pattern

First of all you should create property with employee details in the Employee class:

public class Employee
{
    public Employee()
    {
        EmployeeDetails = new EmployeeDetails();
        EmployeeDetails.EmpID = 123;
        EmployeeDetails.EmpName = "ABC";
    }

    public EmployeeDetails EmployeeDetails { get; set; }
}

If you don't do that, you will create instance of object in Employee constructor and you lose reference to it.

In the XAML you should create instance of Employee class, and after that you can assign it to DataContext.

Your XAML should look like this:

<Window x:Class="SampleApplication.MainWindow"
    xmlns="http://schemas.microsoft.com/winfx/2006/xaml/presentation"
    xmlns:x="http://schemas.microsoft.com/winfx/2006/xaml"
    Title="MainWindow" Height="350" Width="525"
    xmlns:local="clr-namespace:SampleApplication"
   >
    <Window.Resources>
        <local:Employee x:Key="Employee" />
    </Window.Resources>
    <Grid DataContext="{StaticResource Employee}">
        <Grid.RowDefinitions>
            <RowDefinition Height="Auto" />
            <RowDefinition Height="Auto" />
        </Grid.RowDefinitions>
        <Grid.ColumnDefinitions>
            <ColumnDefinition Width="Auto" />
            <ColumnDefinition Width="200" />
        </Grid.ColumnDefinitions>

        <Label Grid.Row="0" Grid.Column="0" Content="ID:"/>
        <Label Grid.Row="1" Grid.Column="0" Content="Name:"/>
        <TextBox Grid.Column="1" Grid.Row="0" Margin="3" Text="{Binding EmployeeDetails.EmpID}" />
        <TextBox Grid.Column="1" Grid.Row="1" Margin="3" Text="{Binding EmployeeDetails.EmpName}" />
    </Grid>
</Window>

Now, after you created property with employee details you should binding by using this property:

Text="{Binding EmployeeDetails.EmpID}"