Analyzing ECB's and CBC's resistence to a known-plaintext attack
I understand that, for a block cipher that receives a key of k bits and processes blocks of n bits (n>k), on ECB mode, Oscar would find false keys and because of that he would need to check more than one pair (i.e. two or three for DES).


That doesn't sound right; if you are assuming that the attacker (Oscar) has a plaintext block $P$ and a ciphertext block $C$, and searches for all keys $K$ to find one that satisfies $C = E_K(P)$, well, if the size of the key space $k$ is smaller than the block size $n$, then one would expect that there would be at most one key $K$ that satifies this relationship.  That is, if we assume that $E$ acts independently with different keys, then $E_K(P)$ is effectively a set of independent variables based on the various values of $K$; since $K$ can take on $2^k$ different values, $E_K(P)$ can take on up to $2^n$ different values.  We're looking for a value $K$ for which $E_K(P)$ is a particular value (C); we know that there is one value $K$ (the correct one) for which this is true; the probability that there is a second (false) key for which this is true is at most $(2^k-1)/2^n$, which is fairly small if $k$ is larger than $n$.

However, on to your real question:


  But how would this attack work in a block cipher that uses CBC mode?


Pretty much the exact same way as it does in ECB; CBC mode is defined as:

$C_i = E_K( P_i \oplus C_{i-1} )$

(where $C_{-1}$ is taken to be the IV)

When the attacker has a known plaintext (that is, he has a plaintext message and its corresponding ciphertext), then the attacker knows the values $C_i$, $P_i$, and $C_{i-1}$, and he can look for a key $K$ that satisfies the value relation exactly as he would with ECB mode.  In addition, if he happens to find two keys $K^1$, $K^2$ that encrypts that block in the exact same way, then he can handle it just like in ECB mode, he steps to another block and does a test encryption (or decryption; whichever is most convienent) on that as well.


  I suppose Oscar still needs to deal with false keys, but now a encryption operation over block i needs the knowledge of block i−1 and possibly the knowledge of IV.


We always assume that the attacker sees the entire ciphertext; hence he sees ciphertext block $i-1$.  It's possible that he doesn't know the IV (we typically send the IV with the message, however that's not a requirement); if he doesn't, all the attacker does is not use the first block for his brute force attack.