The problem statement is in the title. $A_n, B_n$ are random sequences, $C_n$ is deterministic.
The statement itself is complicated because of various notations. But I think the proof relies on this basic question. I'm not familiar with the proof technique when there is a $o(1)$ in side the probability statement and working with $\limsup$.
I know that in the event ${A_n \le B_n + C_n}$, since $A_n > u$ implies $B_n + C_n > u$, the event ${A_n > u}$ is a subset of ${B_n + C_n > u}$. However I feel wrong to write $P(A_n > u) - P(B_n + C_n > u) \le 0$, because it replies on the event ${A_n \le B_n + C_n}$, which has probability goes to 1 as $n\to \infty$. I can't write $P(A_n > u) - P(B_n + C_n > u) \le 0$ with probability converges to 1 after all. That is my starting point, and sooner I'm not sure also how to pull the $C_n$ out and utilize $C_n = o(1)$. If I can manage to this, I still need some help on how to get in touch with the limsup in the target statement. Thank you for all the advice.
In general this is false. Take the case of $A_n = B_n + c_n$ w.p. $1,$ where $c_n > 0$. Suppose that $B_n$ is uniform over $(-c_n/2, c_n/2)$. Then we have $$ P(A_n > 0) - P(B_n>0) = P(B_n > -c_n) - P(B_n > 0) = 1 - 1/2 = 1/2,$$ so the limit in question is positive at $u = 0$. This holds independently of the behaviour of $c_n,$ and in particular also works when $c_n \to 0$ with $n$.
Notice though that when $c_n \to 0,$ the above law on $B_n$ degenerates with $n$ in the sense that $B_n$ converges to a point mass at $0$. This style of counterexample continues to work if $B_n$ goes to some sort of discrete set almost surely. So it is possible that the bound does hold if we demand some smoothness from limiting behaviour of the $B_n$.
Indeed, suppose that $B_n$ satisfies that for any sequence $\delta_n > 0$ such that $\delta_n \to 0,$ and any $u$, it holds that $$ \lim {P}(B_n \in (u-\delta_n, u]) = 0. \tag{*}$$ Then I claim that the bound does hold. To see this, first observe that $$ P(A_n > u) = P(A_n > B_n + c_n, A_n > u) + P(A_n \le B_n + c_n, A_n > u),\\ P(B_n>u) = P(A_n > B_n + c_n, B_n>u) + P(A_n \le B_n + c_n, B_n > u).$$
Now, first consider $$ \zeta_n := P(A_n > B_n + c_n, A_n > u) - P(A_n > B_n + c_n, B_n > u).$$ Clearly we have $$ \zeta_n \le P(A_n > B_n + c_n), \zeta_n \ge -P(A_n > B_n +c _n),$$ and so $\zeta_n \to 0.$ So, we must have that $$\limsup P(A_n > u) - P(B_n > u)\\ = \limsup P( A_n \le B_n + c_n, A_n > u) - P(A_n \le B_n + c_n, B_n > u). $$ To make notation a little more succinct, set $D_n = \min(A_n, B_n + c_n).$ It suffices to analyse $P(D_n > u) - P(B_n > u)$ (why?).
Now, we can break this as \begin{align} &P(D_n > u) - P(B_n > u)\\ &=\, P(D_n > B_n, D_n > u) + P(D_n \le B_n, D_n > u) \\ &\quad - P(D_n > B_n, B_n > u) - P(D_n \le B_n, B_n > u).\end{align} But $D_n\le B_n, D_n > u \implies B_n > u,$ and so we find that for every $n,$ $$ \xi_n := P(D_n \le B_n, D_n > u) - P(D_n \le B_n, B_n > u) \le 0.$$
Further, observe that \begin{align} P(D_n > B_n, D_n > u) - P(D_n > B_n, B_n > u) &= P(D_n > B_n, D_n> u, B_n \le u) \\ &\le P(B_n \le u, B_n + c_n > u) \\ &\le P(B_n \in (u - |c_n|, u]) ,\end{align} which goes to $0$ by our condition $(*)$.
Putting everything together, we can write the bound $$ P(A_n > u) - P(B_n > u) \le P(A_n > B_n + c_n) + \xi_n + P(B_n \in (u-|c_n|, u]).$$ Here the first and last term go to $0$, and the second term is nonpositive, so the $\limsup$ of this term can be at most $0$.
Finally, just to make the point explicitly, note that the condition $(*)$ is also necessary at this level of generality, in the sense that if there is some $u_0$ and $\{\delta_n\}$ such that the limit in $(*)$ is nonzero, then a counterexample exists: taking $A_n = B_n + \delta_n,$ we have that \begin{align} P(A_n > u_0) - P(B_n > u_0) &= P(B_n > u_0 - \delta_n) -P(B_n > u_0) \\ &= P(B_n \in (u_0 - \delta_n, u_0]),\end{align} which converges to something positive. Of course, it would also be interesting to figure out what types of $A_n$ suffice for the result if $B_n$ has this property, but I'm not sure that has a very clean answer.