Consider these four circuits:
simulate this circuit – Schematic created using CircuitLab
How much current is flowing through BAT1 and SW1? The circuit is open so no current can flow. \$0A\$.
How much current is flowing through BAT2 and R1? By Ohm's law: \$9V/100000000\Omega = 0.00000009A\$. That's so close to 0A it like an open switch.
How much current is flowing through BAT3 and SW2? There's no resistance to limit the current, so it is unlimited. \$\infty A\$ (through in practice this can't happen, we are talking about ideals)
How much current is flowing through BAT4 and R2? Again by Ohm's law: \$9V / 0.001 = 9000A\$. That's so much current it's like infinite current, like a closed switch.
An ideal open switch is equivalent to an infinitely large resistor, \$\infty \Omega\$. An ideal closed switch is equivalent to a resistor with no resistance, \$0\Omega\$.
So while a BJT in saturation has some resistance, the resistance is small enough that we can usually consider it to be like a closed switch. Also, though you don't mention it, a BJT in cutoff has some small leakage current, that is, its resistance is very large, but not infinite. Still, we can usually consider it to be like an open switch.
You have to work the transistor in saturation or cutoff mode because if you don't, you get a transistor that's somewhere in between, like a resistor, but not a switch. Instead of almost \$0\Omega\$ or almost \$\infty\Omega\$ you get a number in between. There are many applications of transistors like this (amplifiers), but an amplifier is not like a switch.