Let $f_n$ be integrable functions over $[a,b]$ such that $f_n \to f$ uniformly, then $f$ is integrable over $[a,b]$ and $\lim_n \int_a^b f_n = \int_a^b f$
Proof: We will prove this by definition, let $\epsilon > 0$ then by integrability of $f_n$, exists partition such that if $\lambda(P) < \delta$ then $U(f,P)-L(f,P) < \epsilon /2$
Then since $f_n \to f$ uniformly, then exists $N$ such that for all $n>N$ we have that $|f_n(x)-f(x)| < \frac{\epsilon}{4(b-a)}$ for any choice of point $x \in [a,b]$
It follows that for any two points $s,t \in [a,b]$ that $$f_n(s)-f_n(t)-\frac{\epsilon}{2(b-a)} \le f(s)-f(t) \le f_n(s)-f_n(t)+\frac{\epsilon}{2(b-a)}$$ Denoting $M_i^f-m_i^f=\sup_{s,t\in \triangle x_i}f(s)-f(t)$ then $$M_i^{f_n}-m_i^{f^n}-\frac{\epsilon}{2(b-a)} \le M^f_i-m_i^f \le M_i^{f_n}-m_i^{f^n}+\frac{\epsilon}{2(b-a)}$$ in every subinterval in $P$.
It follows that if we sum the $|P|$ inequalities that
$$\underbrace{\sum M_i^{f_n}-m_i^{f^n}\triangle x_i}_{\epsilon\2}-\underbrace{\sum \frac{\epsilon}{2(b-a)}}_{\epsilon\2} \le \sum M^f_i-m_i^f\triangle x_i \le \underbrace{\sum M_i^{f_n}-m_i^{f^n}\triangle x_i}_{\epsilon\2}+\underbrace{\sum \frac{\epsilon}{2(b-a)}}_{\epsilon\2}$$
The proof is pretty straightforward, a bit technical but that's ok, my question is why are we not changing the equation after denoting $M_i^f-m_i^f=\sup_{s,t\in \triangle x_i}f(s)+f(t)$, it's not necessary that $M_i^f-m_i^f\le M_i^{f_n}-m_i^{f_n}$ because $f_n$ could be constant in $\triangle x_i$ but $f$ not while still being in $2\epsilon$ distance, so why does it work?
The inequality $$ M_i^f-m_i^f\le M_i^{f_n}-m_i^{f_n} $$ is not true and your proof did not use it. What you really use is $$ M_i^f-m_i^f\le M_i^{f_n}-m_i^{f_n}+\frac{\varepsilon}{2(b-a)}. $$ And this inequality is derived from $$ f_n(s)-f_n(t)-\frac{\epsilon}{2(b-a)} \le f(s)-f(t) \le f_n(s)-f_n(t)+\frac{\epsilon}{2(b-a)} $$ (Use the fact that $s$ and $t$ are arbitrary).