To answer this, consider the simpler to understand SCR instead of a triac. A triac is sortof two SCRs back to back and therefore can pass current in both directions. A SCR only works one way but has the same issue of holding current.
Here is a equivalent circuit of a SCR:
SCRs are actually built as one integrated device, but you can conceptualize them as two transistors like this. In fact, you can even make a scr from a NPN and PNP transistor like this if you just want to experiment.
Look at this circuit carefully and see how it works. If somehow a little current were to flow thru one of the transistors, let's say Q1, that causes base current to flow thru Q2, which causes even larger base current thru Q1, which then turns on Q2 even more, etc. Once a little current starts flowing, this circuit latches on.
Now imagine current is flowing and the gate is left open. As long as the current continues, the circuit acts like a switch in the on state. However, below some level of current, the cascading amplifying effect can't be sustained anymore, and the circuit switches off. This minimum level of current so that the device is guaranteed to stay on is the minimum holding current.
This circuit only works with current flowing in one direction whereas triacs work in both directions, but the concept of the minimum current to keep the device on is the same.
The ground connection on the left should not be be there. The whole point of the optoisolator is to provide galvanic isolation between the low voltage and mains sides.
It's also conventional to have signal flow from left to right, so normally the schematic should show the switch at the left and the output on the right. That's a style thing, it does not affect the actual circuit.
The input circuit has a problem- you will destroy the LED as soon as the switch is depressed since there is no series current-limiting resistor.
The output circuit has problems as well, the MOC3042 is not designed to switch a lamp load, it's main purpose is to switch a larger triac. If the incandescent bulb in your schematic is 100W and you have calculated the 580 ohms as the resistance, that is the resistance when it is hot. When you first apply power, the resistance will be much less, perhaps 1/10 to 1/20 of the resistance at operating temperature. That means that the current at switch-on could be in the 5-10A range, which could be enough to damage the device (the only hint is that the repetitive current is limited to 1A in the datasheet, and that will certainly be exceeded at times with such a load).
Even if it survives the initial turn-on, the power dissipation rating will be exceeded. See "Figure 2 On State Characteristics", and the maximum dissipation rating Pd of 150mW - 1.76mW/°C- which implies you should keep current to maybe 25mA to allow for moderately high ambient temperatures.
A small triac (8-16A range) will provide a lot more beefy ratings for switching a serious load, and the MOC3042 can be used to switch it. And don't forget the input resistor.
Best Answer
Your electric fan speed regulator will have to be controlled by something, what will it be? Because I doubt the heater has something like that inside. Also, a gate resistor will only limit the command current but not the current through the device itself.
The way I see it, if you want to control the heating resistance's current (temperature) with a triac you'd have to resort to a dimmer-like circuit:
Do note, though, that you're dealing with \$ \frac{2kW}{230V_{RMS}} \approx 8.7A_{RMS} \$, which is quite a bit higher than your average light-bulb.
As for the heat-sink, it may well be the case. The dissipated power would be:
\$ P_d = V_{t0} I_{T_{AVG}} + R_d I_{T_{RMS}}^2 \$
With the catalog values and the previously calculated one:
\$ Pd = 0.85V \cdot 8.7A*\frac{2\sqrt{2}}{\pi} + 10m\Omega \cdot 8.7^2 = 7.41W \$
\$ T_j = P_d R_{th_{j-a}}+T_a = 7.41W\cdot 50\frac{^\circ C}{W} + 25^\circ C = 395.5^\circ C > T_{j_{MAX}}=125^\circ C \$
I assumed \$25^\circ C\$ ambient temperature, which means you'll need a heat-sink. After a quick math (involving some graphs), you'll need an \$ R_{th_{h-a}} \approx 13\frac{^\circ C}{W}\$ , which would be an aluminium square of 7.5cm (~3in) with a thickness of 1mm. Of course, conditions vary and this is an informative note, only. What thermal contact you're using, how you're mounting the heat-sink, etc, have an influence on the final result.