Why does DESX break if one removed post whitening?
The goal of DESX is to improve the security above the 56-bit level offered by DES. So an attack that's merely as expensive as brute-forcing the DES key is an attack, even if it's still rather expensive.

Define $E_{K,W}^\prime(P) = E_{K}(P\oplus W)$ and $D^\prime$ as its inverse. Then we can relate the two known pairs $(K_1,P_1)$ and $(K_2,P_2)$:

$$\begin{eqnarray} 
P_1\oplus P_2 &amp;=&amp; D^\prime(C_1)\oplus D^\prime(C_2) 
             \\ &amp;=&amp; (D_K(C_1)\oplus W)\oplus (D_K(C_2)\oplus W)
             \\ &amp;=&amp; D_K(C_1) \oplus D_K(C_2)
\end{eqnarray} $$

Since $P_1\oplus P_2 = D_K(C_1) \oplus D_K(C_2)$ only involves $K$ and not $W$ we can simply brute-force that 56-bit key, ignoring $W$, showing that using only pre-whitening does not improve security compared to plain DES. A similar argument can be applied to post-whitening. Note that this attack does not rely on the particular structure of DES (except having a key short enough to brute-force) and can be applied to any other blockcipher.