Differential pair biasing by current sources
There ar two points to be considered:

At first, in reality there are no real current sources. However, we are using such a symbol to indicate that there is a voltage source with a comparatively very large source resistance (either a large ohmic resistor or - very often - a transistor based cicuit with a very large dynamic output resistance).

Secondly, the base nodes must not be open. The whole circuit can be designed using split supply with an additional negative supply voltage. In this case, the base nodes need a DC connection to ground. Or - as in your case - we need a suitable dc voltage at the base (0.7V about the common emitter potential) - created with a suitable voltage divider.


EDIT (Addendum): To Keno.
Perhaps the following helps to UNDESTAND the circuit.

We want to build a differential amplifier.

No problem: The BJT can amplify the voltage difference between both input nodes: Base and Emitter: Ic=f(Vbe)

Problem: Both input resistances are very different (input resistance at the emitter very low).

Solution: We are using an additional high-impedance common collector stage (transistor T1, emitter follower) in front of the original stage with the emitter input. Now THIS stage (transistor T2, emitter input) works as a common base stage.

Result: Two stages in series: Common collector (T1) and common base (T2) with one single common emitter resistor.

Final consideration: The whole circuit is symmetrical (when we include a collector resistor into the collector path of T1 - without changing its property working as a emitter follower). Hence, we can see and describe the operation of the whole circuit from the left or from the right. In both cases, series connection of two stages (emitter follower and common base).

Now we can operate this two-transistor circuit from either side - or at the same time from both sides (superposition principle) -