Encrypting the same message using different schemes $E_1$ and $E_2$ are IND-CPA secure encryption schemes.
$E$ is defined as: 
$k_1,k_2 \leftarrow K_1 \times K_2$ .
$E_{k_1,k_2}(m) \leftarrow E_{1,k_1}(m)||E_{2,k_2}(m)$.

Hope the notations are in an obvious manner.
$\leftarrow$ means randomly choosing an element from the corresponding distribution. || stands for concatenation.

My question is whether the encryption scheme $E$ is IND-CPA secure.

Intuitively, this would be secure. If you can not get any information from either $E_1$ or $E_2$, should you get no information given both. (Of course there is a loophole here: $E_1$ or $E_2$ by itself may not leak information, however the combination of them may give some information away.)

When I tried to prove it I can not reduce the break of  $E$ to either $E_1$ or $E_2$. Neither can I find a counter example.
Yes, $E$ will be always be secure.  This follows from a standard type of proof called a hybrid argument.  Giving the full details would be tedious, so here is a sketch in case you are familiar with hybrid arguments:

We define games $H_0,H_1,H_2$.
We let $H_0$ be the IND-CPA game, but with the game's secret bit hardcoded to $0$.  So the game always outputs $E_{1,k_1}(m_0)\|E_{2,k_2}(m_0)$ when the adversary submits $(m_0,m_1)$ as its challenge. 

Next define $H_1$ to always return $E_{1,k_1}(m_1)\|E_{2,k_2}(m_0)$ always, and finally $H_2$ to always return $E_{1,k_1}(m_1)\|E_{2,k_2}(m_1)$.

A successful IND-CPA adversary $\mathcal{A}$ against $E$ is one that distinguishes $H_0$ and $H_2$.  The games $H_0,H_1,H_2$ are chosen so that we can use $\mathcal{A}$ to build adversaries $\mathcal{A}_0, \mathcal{A}_1$ to distinguish between $H_0$ and $H_1$ or between $H_1$ and $H_2$. But this will mean that at least one of the adversaries breaks the IND-CPA security of $E_1$ or $E_2$ respectively.