Attempt to use 2n2222 as a switch, but it seems to work below 5v (all the way down to < 1v). Why?
I guess I don't want you to lose sight of the fact that a voltage does, in fact, control the collector current of a BJT. You are right to hold that idea. But let's look at the simplified case (active, not saturated):

$$I_C=I_{SAT}\cdot\left(e^{\frac{V_{BE}}{V_T}}-1\right)$$

(\$I_{SAT}\$ is a model parameter constant. You don't need to know its value, but for small signal BJTs it is often around \$I_{SAT}\approx 2\times 10^{-14}\:\textrm{A}\$. Also, \$V_T\$ is the thermal voltage and at room temperature is about \$26\:\textrm{mV}\$.)

You can see that the base current doesn't even show up in the equation. So the base-emitter voltage does control the behavior in the active region. However, a BJT also needs some current supplied to the base to compensate for "recombination" that occurs as a "side-effect." It turns out that for a specific BJT you have in your hand, and over a relatively wide range of collector currents, that the amount of base recombination current is a relatively fixed proportion of the collector current. (This proportion is either \$\beta\$ or \$\frac{1}{\beta}\$, depending on which end of things you are looking at.)

What is NOT true is that if you pick up a second BJT from the same pile, that the "relatively fixed proportion" (\$\beta\$) will be the same value as in the first BJT. BJTs themselves vary as to the exact proportion. However, for a given family their \$\beta\$ will be within a factor of 2, or so, of each other. So if the datasheet says \$\beta=200\$ then you probably can figure it will be more than \$\beta=100\$ and probably less than \$\beta=400\$. This is why people will constantly hammer in the idea that you cannot have a circuit design that depends on a precise value for this constant. However, it's still true that for a given part, it remains surprisingly fixed in value over several orders of magnitude of collector current. (From about the minimum useful collector current to the point where current crowding and wire bonding resistance becomes a problem at higher currents.)

It can be proved using the above equation that the collector current can be 10 times more, or 10 times less, and that the base-emitter voltage only shifts by about \$60\:\textrm{mV}\$ (at room temp, anyway.) So this also means that the base-emitter voltage is surprisingly stable as the collector current changes a lot.

The problem in operating a BJT by controlling the base voltage is exactly because of this last point and because of the variability between BJTs. Suppose you know that if \$V_{BE}=700\:\textrm{mV}\$ then \$I_C=5\:\textrm{mA}\$ for BJT #1. You've proved it by testing it. Now, you pick up another one, #2, and apply the same very precise voltage to its base. Well, the collector current might be \$I_C=20\:\textrm{mA}\$ because #2 only needed \$670\:\textrm{mV}\$ for \$I_C=5\:\textrm{mA}\$ and you just supplied it \$30\:\textrm{mV}\$ more than it needed. So it responded with a much higher collector current. This is a problem.

The solution is to arrange things so that you supply adequate current, instead, and just let the BJT "find" its \$V_{BE}\$ on its own. Leave that for the BJT. This is a big reason why many circuits use current control, rather than voltage control, despite what the above equation suggests. Since the base current is a fixed portion of the collector current and this portion is remarkably stable over a wide range of collector currents, this is a reasonable way to go.



Now we can get to your circuit.

A designer would think: 


The LED needs \$20\:\textrm{mA}\$
Using a BJT for a switch, means "saturated" and \$I_C=20\:\textrm{mA}\$
Saturated implies \$\beta\$ that is about a tenth of the "active mode" for the BJT, so \$\beta=10\$ and \$I_B=2\:\textrm{mA}\$ unless the datasheet indicates otherwise
In saturation, assume \$V_{BE}=700\:\textrm{mV}\$ unless the datasheet indicates otherwise.
Compute base resistor as \$\frac{V_{CONTROL}-V_{BE}}{I_B}=\frac{5\:\textrm{V}-700\:\textrm{mV}}{2\:\textrm{mA}}=2150\:\Omega\$


And that's the process.

(Some added things may occur. One might check on the possible dissipation of resistors and the BJT.)

In your case, someone just used a \$1\:\textrm{k}\Omega\$ resistor. All this means is that they supplied even more recombination current supply to the BJT. While it may not be needed, it won't hurt (except if this is a battery application or if it might cause the BJT to dissipate too much.)



EDIT: The equation at the outset above can be reversed to solve for the base-emitter voltage needed for some given collector current:

$$V_{BE}=V_T\cdot\operatorname{ln}\left(1+\frac{I_C}{I_{SAT}}\right)$$

Since the fraction inside the log function is usually very, very much larger than 1, the above equation is usefully shortened to the following, without any meaningful loss of generality:

$$V_{BE}\approx V_T\cdot\operatorname{ln}\left(\frac{I_C}{I_{SAT}}\right)$$

Using that equation, you can figure out the change in \$V_{BE}\$ if you know two different collector currents as:

$$\Delta V_{BE}\approx V_T\cdot\operatorname{ln}\left(\frac{I_{C_2}}{I_{C_1}}\right)$$

This is how I computed an example I gave you earlier, where BJT #2 gave an \$I_C=20\:\textrm{mA}\$ and BJT #1 gave \$I_C=5\:\textrm{mA}\$:

$$\begin{align*}
\Delta V_{BE}&amp;\approx V_T\cdot\operatorname{ln}\left(\frac{I_{C_2}}{I_{C_1}}\right)\\\\
&amp;\approx 26\:\textrm{mV}\cdot\operatorname{ln}\left(\frac{20\:\textrm{mA}}{5\:\textrm{mA}}\right)\\\\
&amp;\approx 36\:\textrm{mV}
\end{align*}$$

The above calculation means several things. It means that if you increased the base-emitter voltage on BJT #1 from \$700\:\textrm{mV}\$ to \$736\:\textrm{mV}\$, you'd increase its collector current from \$I_C=5\:\textrm{mA}\$ to \$I_C=20\:\textrm{mA}\$. But also that if you decreased the base-emitter voltage on BJT #2 from \$700\:\textrm{mV}\$ to \$664\:\textrm{mV}\$, you'd decrease its collector current from \$I_C=20\:\textrm{mA}\$ to \$I_C=5\:\textrm{mA}\$.