Breadboard 7408 2 input AND gate always supplying current, regardless of input
Yup. That's a classic, all right. One of the interesting characteristics of the 7400-series TTL logic family is that an open input behaves like a logic HIGH. It's not good practice to do this, since such unused inputs can act like antennas and start picking up noise, but it's working just fine in your case. 

In your circuit, tie each input to +5 with a 1k resistor. When you want to drive that input LOW, just ground it.

And since you're just starting, you need to start developing good habits. That means do not leave any inputs on a chip floating. You can usually get away with it with TTL and LSTTL, but it will drive you crazy if you do it with CMOS. In this case, tie all 6 unused inputs of the 7408 together, and tie the whole thing to +5 with a 1k resistor.

Also, get a 0.1 uF ceramic capacitor, and connect it between pins 7 and 14. This is called a decoupling, or bypass, capacitor, and for more complicated circuits one capacitor per IC is a very, very good idea, with each capacitor connected right to the power and ground pins of its' IC. It doesn't have to be a high-voltage capacitor, since it will only have 5 volts on it.