Why do we connect pin1 to 5V? is it not the rest pin and should it not be connected to the DTR on the FTDI end? Or is it actually always fed 5V and the DTR actually resets the chip by cutting the current?
Pin 1 is not connected directly to 5V, it's connected through a 10K pullup resistor. The resistor helps to ensure that reset is held high and that no unintended resets occur. It can still be pulled low with a button, or the FTDI interface does it through a 100nF capacitor, this generates enough of a negative-going pulse to reset the MCU.
By the way, the component labeled 10K2 is not a resistor, that is the symbol for a capacitor.
One of the non-obvious things with Eagle at first is that components are not connected with wires! Use NETs instead. If the toolbar is docked on the left side of the window, the NET icon is down a few from WIRE and on the right. So bad news, all the wires need to be changed to NETs. That way, the dots will automatically appear showing connections.
Another nice thing with NETs is that they can be named. When there are two (or more) separate NETs on a schematic, but they are given the same name, then Eagle understands that they are connected together. Vcc and GND are common examples. I'm attaching my basic ATmega328 circuit. Note that pin 1 hooks to a NET named /RESET, and that there is another NET named /RESET above the reset switch. Eagle understands these two to be electrically connected. When using this technique, it's strongly suggested to LABEL the NETs so that it's clear when looking at the drawing that they're connected. In fact, if you don't, I believe Eagle's electrical rule check will complain about it. This is a feature some people love to hate, and while a person can certainly go too far with it, OTOH it does have its place.
One other feature I like to include is a diode on the power input to prevent damage if reverse-polarity power is accidentally connected. Good insurance for a few pennies.