For this proof there is a mistake. Find where it is.
$1. \space \lnot a\rightarrow b \qquad \qquad$ Premise.
$2. \space a\rightarrow c \qquad \qquad \space \space$ Premise.
$3. \lnot b \qquad \qquad \qquad \space$Preimse.
$\boxed {4. \lnot a \qquad \qquad \qquad assumption. \\5. \space b \qquad \qquad \qquad \space \rightarrow elimination \space(1,4). \\6. \space F \qquad \qquad \qquad \space \lnot \space elimination \space (3,5).} $
$7. \space a \qquad \qquad \qquad \space $F elimination $(6)$.
$8. \space c \qquad \qquad \qquad \space \rightarrow$ elimination $(2,7)$.
I said the mistake was on line $7$ because when concluding that the assumption $b$ being false than you can conclude anything you want, and what happened is the person concluded a, but this should still be the assumption box?
There is a mistake in this, and I think it is at $7$, but not sure. Any justification to my logic would be appreciated.
We'd have to know the exact rules of the particular system that you are working with in order to answer your question.
Line 7 makes sense, at least logically: if $\neg a$ leads to a contradiction, then (at least in classical logic) you can conclude $a$.
However, normally you would call this $\neg$ Introduction or $\neg$ Elimination, and you'd point to the whole box (i.e. lines 4-6) to do this. And if the $F$ Elimination is defined as: From $F$ you can infer anything you want (and it looks like that's what is going on, given that the justification only points to line 6), then I am with you: whatever you infer from $F$ should still be within the box.
However, I am at the same time confused by the justification of line 6 ... in my system this would be called $F$ Introduction, rather than $\neg$ Elimination .... so again, it would help to know the exact definition of the rules of your system to really answer your question.