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For a given teleportation process as depicted in the figure, how one can say that teleporting the qubit state $|qrangle$ has not cloned at the end of Bob's measurement?

Firstly, carefully read through the formal presentation of the protocol as described on Wikipedia. Secondly, there's nothing to prove as such here. It is evident from the teleportation protocol itself.

Yes, at the conclusion of the teleportation protocol, Bob's qubit certainly will take on the same state as Alice's qubit's initial state i.e. $|qrangle$. However, Alice's qubit will now exist as an inseparable part of an entangled Bell state. You haven't really been able to create a "copy" of Alice's qubit, in the sense that Alice's qubit's initial state is "destroyed" in the process and is no longer $|qrangle$.

In (crude) analogical terms, in the quantum teleportation protocol you "cut and paste" the state $|qrangle$ rather than "copying and pasting". So there's no violation of the No cloning theorem!

Cloning means the generation of $|qrangle|qrangle$ from $|qrangle|0rangle$. This is not what happens in teleportation. Teleportation is kind of a swap operation, i.e. something like $|qrangle|0rangle to |0rangle|qrangle$.

I.e., if we have the entire initial state is written as follows

$$| q rangle otimes | beta_00 rangle =(alpha | 0rangle+beta | 1 rangle ) otimes frac1sqrt2( |00rangle+| 11 rangle)$$ $$= frac1sqrt2(alpha | 000rangle+alpha | 011 rangle+beta | 100 rangle+beta | 111 rangle ),$$

then, in step three, we obtain the state (the system state collapses to one of four possible result after measurement),

$$|psi rangle equiv 11 rangle frac 1 rangle-beta 2 ,$$

where the right two-qubit is for Alice and the superposed qubits with corresponding probabilities $alpha$ and $beta$ are for Bob

hence we can no longer write

$$| q rangle_A otimes | textsomething rangle_B,$$

namely, Alice's $| q rangle_A$ dissapeared.