Why is only one oracle + challenge-response phase sufficient for IND-CPA security? The CRYPTUTOR wiki from UIUC uses the following definition of the IND-CPA game:


  
  We (privately) choose a key K according to the key generation
  algorithm: K $\gets \mathsf{KeyGen}$.
  We (privately) choose a random bit b $\gets \{0,1\}$.
  Repeatedly do:
  
  
  A is allowed (any number of times) to query an oracle that computes the functionality $\mathsf{Enc}_K$.
  Challenge: A outputs two messages, $m_0\, $and $m_1\,$.
  Response: We give A the ciphertext $\mathsf{Enc}_K(m_b)$
  
  
  
  A outputs $b'\,$ (i.e, a guess for our b).
  
  We say that the advantage of A in this experiment is $\Pr[ b = b' ] - 1/2$. 


Later in the Modifications section it is stated as follows:


  Using an application of a hybrid argument, it can be shown that only one "challenge-response" phase is necessary. It does not affect the definition of security to remove the repeating loop (line 3). 


I am not able to come up with any hybrid arguments to prove this.
EDIT: I realised that I assumed an IND-CPA game where the adversary has pre- and post-challenge access to the encryption oracle and not only pre-challenge access. I'll edit my answer soon.

I'll give the idea and leave the concrete analysis to you. You want to show that single-query IND-CPA implies multi-query IND-CPA or in other words if there is an adversary against multi-query IND-CPA, then there is also one against single-query IND-CPA.

Assume that you have an adversary $\cal A$ that that wins the multi-query IND-CPA experiment with $q$ challenge queries. 

For our argumentation below, we construct an adversary $\cal B$ that uses only one challenge query where $\cal B$ interacts with an single-query IND-CPA challenger $\cal C$ of a symmetric encryption scheme.

The idea is that $\cal B$ chooses an index $i\in [q]$ and proceeds as follows:

Let $\cal B$ initialize the single-query IND-CPA game with a challenger $\cal C$ and:


For all challenge query $(m_0^j,m_1^j)$ with $j&lt;i$ from $\cal A$, $\cal B$ calls the encryption oralce of $\cal C$ with $m_0^j$ and returns the obtained ciphertext to $\cal A$
For the challenge query $(m_0^j,m_1^j)$ with $j=i$, $\cal B$ queries $(m_0^j,m_1^j)$ to the challenge oracle of $\cal C$ and returns whatever $\cal C$ returns
For any challenge query $(m_0^j,m_1^j)$ with $j&gt;i$ from $\cal A$, $\cal B$ calls the encryption oralce of $\cal C$ with $m_1^j$ and returns the obtained ciphertext to $\cal A$


If $\cal A$ outputs its guess $b'$ to $\cal B$, then $\cal B$ forwards $b'$ to $\cal C$ 

Now, define the $j$'th hybrid $H_j$ to be the version of the multi-query IND-CPA game where the first $j$'th queries are with respect to $m_0^j$ (and all remaining with respect to $m_1^j$). It is clear that in $H_0$ all queries are with respect to $m_1$ and in $H_q$ all queries are with respect to $m_0$. Consequently, a distinguisher between these two extreme hybrids $H_0$ and $H_q$ clearly yields a mutli-query IND-CPA adversary. 

If we now assume that there is such an multi-query IND-CPA adversary $\cal A'$ that has non-negligible advantage, then if we move from $H_0$ to $H_q$, there must be an $i, 0\leq i &lt; q$ such that for the neighboring hybrids $H_i$ and $H_{i+1}$ it must hold that their distance must be non-negligible (as the number of queries $q$ is polynomially bounded). Now, you can argue that any distinguisher between two neighboring hybrids yields a distinguisher for the single-query IND-CPA game, by exactly using the adversary $\cal B$ defined above. Finally, you will get that the advantage for an adversary of winning the multi-query IND-CPA experiment is less or equal then $q$ times the advantage of an adversary winning the single-query IND-CPA experiment, i.e., any multi-query adversary with non-negligible advantage yields a single-query adversary with non-negligible advantage (which proves what you want to show).