Problem: The superior limit of a sequence of sets are unaltered if a finite number of terms of the sequence are changed.
My Attempt: Given a sequence of sets $E_1, E_2, E_3....$ denoted by $E_n$ without loss of generality let the first $k$ terms of the sequence be altered so that we have the following sequence $E_1',E_2',...,E_k',....$ which we will denote by $E_n'$. Now we know that $$\lim\sup E_n=\bigcap_{n\geq1}\bigcup_{j\geq n}E_j.$$ We have to show that $$\bigcap_{n\geq1}\bigcup_{j\geq n}E_j=\bigcap_{n\geq1}\bigcup_{j\geq n}E'_j.$$ Let $x\in\bigcap_{n\geq1}\bigcup_{j\geq n}E_j$ then $\forall n\geq 1,\exists j\geq n$ such that $x\in E_j.$ Now if $n\geq k+1$ then we can use the same $j\geq n$ to get that $x\in E'_j=E_j.$ However, if $1\leq n\leq k$ then if the $j\geq k+1$ then we can again use the same $j'=j$ however if $j\leq k$ then we can always find an index bigger than $k$ as we know that $x$ belongs to all the intersections. And thus we can set $j'=j_{\text{new index}}.$ This proves the first inclusion. I guess the proof for the other inclusion works in the same manner.
This is my first encounter with the notion of sequences of sets and so I am not sure whether my proof is correct. I would be grateful if someone could give some feedback.
Globally, it is correct.
I note some doubt from you when you write “I guess the proof for the other inclusion works in the same manner”. Of course it does!
It is much simpler to approach the problem like this:$$\tag{1}x\in\limsup E_n$$means that $x\in E_n$ for infinitely many $n$'s. Therefore, if you change only finitely many $E_n$'s, $(1)$ still holds.