Hi Electricman,
Don't bother with cheap IR headphones. They won't work very well and will contain cheap Chinese parts you won't be able to find out about or find.
Nice try with designing your own IR transmitter but electricity and electronics don't work like that:
1) A little 9V battery is 9V only when brand new and with no load. If you try to get 450mA out of a new 9V alkaline battery, its voltage would be 7V immediately. With a 60mA average load when new, it will be down to 7.2V and 22mA average in 4 or 5 hrs.
2) You can't neglect the voltage drop (not resistance) of the IR LEDs, they need about 1.5V each, or 6V for 4 in series. Check their datasheet for their max average current rating, it might not be as high as 60mA.
3) The absolute maximum current for your BC548 transistor is only 100mA. Select a stronger transistor or FET for 450mA peak.
Let's investigate an IR transmitter design with your circuit. You want 450mA pulses for good range. You can use a big filter cap across the battery to provide the peak current while the battery provides the average current. You operate the 555 with a 60/450 duty-cycle so the LEDs dont burn out:
1) The four 1.5V LEDs need 6V and the stronger transistor will drop 0.2V, leaving 2.8V for the current-limiting resistor.
2) With a new 9V battery, the current-limiting resistor calculates to be 2.8V/450mA = 6.22 ohms.
3) When the battery voltage drops to 7.2V, the LEDs plus transistor will still drop 6.2V, leaving 1V across the current-limiting resistor and therefore the peak current pulses will be down to 161mA.
You can see how the LED current reduces to nearly 1/3 as the battery voltage runs down. Maybe there is a way to increase the duty-cycle as the battery voltage reduces, but then the battery would last only 3 hrs.