Setting the binding source of a control templates element

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C#

I have the following control template;

<ControlTemplate x:Key="NationalTweetLabel" TargetType="{x:Type Label}">
            <Grid Width="447" Height="129" RenderTransformOrigin="0.5,0.5">
                <Grid.RenderTransform>
                    <TransformGroup>
                        <ScaleTransform ScaleX="1" ScaleY="1"/>
                        <SkewTransform AngleX="0" AngleY="0"/>
                        <RotateTransform Angle="0"/>
                        <TranslateTransform X="0" Y="0"/>
                    </TransformGroup>
                </Grid.RenderTransform>
                <Border Margin="0,0,0,31" VerticalAlignment="Stretch" Width="447" Height="97" Opacity="1" Background="#FF00FF18">
                    <TextBlock HorizontalAlignment="Left" VerticalAlignment="Top" Width="448" Height="72.5" Text="{Binding Message}" TextWrapping="Wrap"/>
                </Border>
                <TextBlock Name="Tweet" HorizontalAlignment="Right" Margin="0,0,45,8" VerticalAlignment="Bottom" Text="{Binding User}" TextWrapping="Wrap" d:LayoutOverrides="Height" RenderTransformOrigin="6.98,1.161"/>
                <Image HorizontalAlignment="Right" Margin="0,0,-9.5,-6" VerticalAlignment="Bottom" Width="51" Height="44" d:LayoutOverrides="Height" Source="bluebird.png"/>
                <Image Margin="0,0,0,31.5" Source="Images\top-message-layer.png"/>
            </Grid>
        </ControlTemplate>

The text property of the textblock "Tweet" is set to a binding source, which I want to be a string called User. I create a label in code and apply the template to it;

TextBlock textBlock = new TextBlock();
            textBlock.Template =(ControlTemplate)FindResource("NationalTweetLabel");

I don't know how to set the source of the binding for the Tweet textblock. I found code to get the element of the control template using textBlock.Template.FindName, but I don't know what to do with it after. Any help will be appreciated. Let me know if I haven't explained clearly enough.