Jgord has already covered the main problem of the reset being tied low (reset is active low so it needs to be held high during operation), so I'll just add a couple of points (main point on capacitors at the bottom if you want to skip the boring stuff ;-) ):
You say the input voltage to the 5V regulator is ~5.47V and the output is ~4V. This means the regulator is not regulating.
This is because a linear regulator needs an input voltage somwhat higher than the output voltage to maintain regulation (i.e hold it's output at 5V) How much higher this needs to be is referred to as the "dropout voltage". There are two main types of linear regulator - "Standard" and "LDO" - the basic difference is that the LDO use an open collector rather than emitter follower used in a "standard" regulator, which enables a lower dropout voltage.
Anyway, if we look at the regulator used in the Arduino UNO (NCP1117 and MC33269 in different revisions, both very similar), it is an LDO type with a dropout voltage of around 1.4V max (MCP33269 1.2V).
This means the input voltage has to be the regulation voltage (5V in this case) plus 1.4V to maintain the regulation, so at least 6.4V is needed. If we look at the Arduino specs we see a recommended range of 7V-12V for the input voltage (to leave a bit of headroom is always a good idea)
The upper limit is dictated by the maximum input voltage of the regulator or the power dissipation, which ever "comes first". In this case it is the power dissipation. If we assume the Arduino is capable of drawing 200mA at full tilt, then with 12V input the regulator has to dissipate (12V - 5V) * 0.2A = 1.4W.
This doesn't sound like a lot of power to dissipate, but the package is a SOT223 with a thermal resistance from junction to ambient (θja) of ~160°C/W. This means at 1W the package will rise by 160°C above ambient, so at 1.4W it will rise by 1.4 * 160 = 224°C. So this is clearly not possible.
If we assume a maximum ambient of 50°C, then the maximum current we can draw through the regulator with an input voltage of 12V will be:
(150°C - 50°C) / 160°C/W = 0.625W is the maximum we can dissipate so:
0.625W / (12V - 5V) = 89mA is the maximum we can draw.
At 7V, the situation wouldn't be as bad:
0.625W / (7V - 5V) = 312mA
So you can see why they advise to limit the voltage to 12V maximum.
You need some capacitors on your regulator input and output:
The output capacitor is required for stability, and the input capacitor is a good idea for transient response (may be required for stability if the regulator is a more than a few inches from the power source - see datasheet page 8, linked to above)
Microcontroller bypass capacitor
Finally, you also need a bypass capacitor on your microcontroller Vcc and AVcc pins, a typical value is 100nF or 1uF ceramic (from each pin to ground, place as near the pins as possible) Have a look in the datasheet for recommended decoupling and follow the example circuit (should be one provided)
If you are using the analogue features then some isolation of Vcc and AVcc is a good idea (usually something like a series 10Ω resistor and/or inductor is recommended between the Vcc and AVcc supplies, again the datasheet should have details)