The proof you've shown proves that -for a given bit of the string, $\Delta m_1$- we know that the tension acting to the right, T1 must equal the tension to the left, T2 (Due to Newton's SECOND law).
Now, consider the piece of string directly the right of the first peice we were considering. Call it $\Delta m_2$. This is actually the piece of mass responsible for the force T1 onto the first mass $\Delta m_1$. Since T1 puts a force to the right on mass $\Delta m_1$ there must be an equal and opposite force acting from $\Delta m_1$ onto $\Delta m_2$, call this force T3. So we know that T3 acts to the LEFT onto $\Delta m_2$ and also that T3=T1 (Due to Newton's THIRD law).
Finally, we know there is a force T4 which acts to the RIGHT on $\Delta m_2$ and that (by the same Newton's SECOND law argument as above) must equal to T3 for a massless string.
So in summary:
T1 = T2 (Newton's Second law, your argument above)
T3 = T2 (Newton's Third law between the two elements of the string)
T3 = T4 (Newton's Second law applying your argument above to the second piece of string)
So you can go along this string the whole way and you will see that each element of the string has a force T acting to the left and force T acting to the right. You will see that the force to the left always equals the force to the right but furthermore, you will see that T is the same all throughout the string.
A little bit shorter proof by contradiction: Say the tension is T1 in one part of the string and T2 in another part. Well then there must be a piece of the string in between those two sections of string where the tensions acting to the left and right are imbalanced. The piece of string would be a contradiction to the argument you've given above that all pieces of string must have balanced forces on them to satisfy Newton's second law.
