You have the answer there in the properties page. The drive current is 10mA. With your 10k resistor, how much current do you think is going through the LED?
The forward voltage of the LED is (almost) constant (is does vary with current/temperature slightly), so changing the resistor value won't change the voltage over the LED. What it will do, is change the current going through it.
Your properties page shows a 2.2V drop over the LED. You have a source voltage of 5V. To find the current through the LED, use the formula: Iled = Vs-Vled/R
Using this, you can see the current going through your LED is: (5-2.2)/10000 = 280uA which is nowhere near enough to turn it on.
Try changing your resistor to something like 220 or 330 ohms and you will see it turn on.
Here is your circuit as you have it set up (using the exact same components):
Look at the current. Not enough to light it up.
Here it is with the resistor changed:
It was pointed out by brhans that I may have misread the question. If you are wondering why the voltage you are measuring is not the 2.2V you are expecting, it is because you are not measuring across the LED. You are actually measuring the voltage across the switch. This means anything you change above it will not affect the voltage across it, as the switch is constant. If you want to measure the voltage over the LED, then you need to move your probe above it. With the switch open, it will be near enough 5V, and the bottom probe will be the voltage you are measuring, which is the diode drop with zero current flowing. Close the switch and you will see the voltage changes. To show you what I mean:
This shows the expected results. Close the switch and you will see the numbers change:
Note that the switch has an open resistance of 100Mohms. As andy aka points out, this will account for the voltage level you see.