Problem in proof of Chinese remainder theorem, and applying it. Please don't mark it as duplicate. First read the whole question. 
So Chinese Remainder Theorem states that,:

Let $n_1,n_2,...,n_k$ be $k$ positive integers which are pairwise relatively prime. If $a_1,a_2,...,a_k$ are such that $(a_j,n_j)=1$ for $j=1,2,...k$ then the congruences $$a_1x \equiv b_1(\mod n_1),a_2x \equiv b_2(\mod n_2),...,a_kx \equiv b_k(\mod n_k)$$ 
have a common solution which is unique modulo $[n_1,n_2,...n_k]$. 
PROOF: Consider $a_jx \equiv b_j(\mod n_j)$. Since, $(a_j,n_j)=1$, we always have a solution for $a_jx \equiv b_j(\mod n_j)$ whatever be $b_j$. $(1)$ 
Choose a solution $C_j$ for $a_jx \equiv b_j(\mod n_j)$ for $j=1,2,...,k$. We have $[n_1,...,n_k]=n_1..n_k$ since they all are co-prime. Call this number $M$. If $m_j=\frac M{n_j}$ we see that $(m_j,n_j)=1$ Solving $m_jx\equiv 1(\mod n_j)$ we have a unique solution $x\equiv m_j'(\mod n_j)$. $(2)$ 
Wherever I have marked a number $(1)$ or $(2)$, I didn't understand the step. 
Also, I didn't understand the steps that are taken from now onwards. 
This gives $m_jm_j' \equiv 1(\mod n_j)$. Take $x_0=c_1m_1m_1'+c_2m_2m_2'+...+c_km_km_k'.$ For $i\neq j$, $n_i$ divides $m_j=\frac{n_1n_2...n_k}{n_j}$. Therefore $$a_jx_0=\sum\limits_{i=1}^{k}a_ic_im_im_i'\equiv a_jc_jm_jm_j' (\mod n_j)$$ 
$$\equiv a_jc_j(\mod n_j)$$ since $m_jm_j' \equiv 1(\mod n_j)$ 
$$\equiv b_j(\mod n_j)$$ for $j=1,2,...,k$. 
Thus, $x_0$ is a common solution to our system of congruences. If $x$ is any other solution of the same system then $x_0 \equiv c_j \equiv x(\mod n_j)$. This means that $x_0-x$ is a common multiple of $n_1,n_2,...,n_k$ and hence $x_0-x$ is a multiple of $[n_1,n_2,...,n_k]=M$. Therefore $x\equiv x_0(\mod [n_1,...,n_k])$ 

Now what does the writer mean by $m_j'$? Where did the $'$ come from? 

Also, how to apply it, like in this example: There are $x$ eggs in a basket. If counted in pairs, $1$ remains. 
If counted in groups of three, $2$ remain. 
If in groups of four, $3$ remain. 
If in groups of five, $4$ remain. 
If in groups of six, $5$ remain. 
If in groups of seven, $0$ remain. So find $x$.

I made the congruences easily, but how to use CRT here?


Here 

we have a unique solution $x\equiv m_j(\mod n_j)$. $(2)$ 

it should say instead $x\equiv m_j'(\mod n_j)$. The $m_j'$ is the solution of $m_jx\equiv 1(\mod n_j)$. To use the notation $m_j'$ is somewhat common, but it is not crucial either. You could just as well say let $u_j$ be the solution of  $m_jx\equiv 1(\mod n_j)$, and then write $u_j$ wherever you have $m_j'$. 
For you specific question to apply CRT in the form you quote there, you need to work on the congruences a bit to have  co-prime moduli, or you use a version of CRT that allows for non co-prime moduli.  
