Which is wrong in the proof? Consider the following steps:
\begin{align}
     a &= x \\
   a+a &= a+x    && \text{[add }a\text{ to both sides]}\\
    2a &= a+x    && \text{[}a+a = 2a\text{]}\\
 2a-2x &= a+x-2x && \text{[subtract }2x\text{ from both sides]}\\
2(a-x) &= a+x-2x && \text{[}2a-2x = 2(a-x)\text{]}\\
2(a-x) &= a-x    && \text{[}x-2x = -x\text{]}\\
     2 &= 1      && \text{[divide both sides by }a-x\text{]}
\end{align}
Which step is wrong in this proof?

$$\begin{align}
a & = x \\
a + a & = a + x \\ 
2a & = a+x \\
2a - 2x & = a + x - 2x\\ 
2 (a -x) &= a-x 
\end{align}$$
All is OK so far but in the next step you are trying to divide $0$ by $0$ which is not allowed $\frac{0}{0}$ is undefined.
Note: $a = x$ so $a - x = 0$
