Prove that if the sequence of increasing functions $(f_n)$ converges to $f$, then $f$ is increasing 
Let $a,b\in \Bbb R$, $a<b$ and for $n=1,2,3,\ldots$ let $f_n\colon [a,b]\to \Bbb R$ be an increasing function (i.e. $f_n(x)\leq f_n(y)$ if $x\leq y$). Prove that if the sequence $f_1,f_2,f_3,\ldots$ converges to $f$ then $f$ is increasing, and that if $f$ is continuous then the convergence is uniform.


Let $(f_n)$ be a sequence of increasing functions from $[a,b]$ to $\mathbb{R}$ and $f$ be its pointwise limit.
Choose $x$ and $y$ such that $x\ge y$. There are $\epsilon>0$ and $N,M$ such that $|f_n(x)-f(x)|<\epsilon$ and $|f_m(y)-f(y)|<\epsilon$ whenever $n>N$ and $m>M$. If we choose $k>\min\{N,M\}$  then we have
$$f_k(x)-f(x)<\epsilon$$
$$-f_k(y)+f(y)<\epsilon$$
thus $0\le f_k(x)-f_k(y)< f(x)-f(y)+2\epsilon$. Since we can make $\epsilon$ arbitrarly small, $f(x)\ge f(y)$.
I find it difficult to prove the second part. Any hint please?

Suppose the contrary:
$$\exists \varepsilon > 0 : \forall n \in \mathbb{N}, \exists x=x(n) \in [a,b] , m \ge n : |f_m(x)-f(x)| \ge \varepsilon$$
Note that $x(n)$ must have converging subsequence $y_n \to y$ and due to $f$ being continuous, $f(y_n)$ also converges to $f(y)$. Finally:
$$f(y) = \lim_{n \to \infty} f_n(y)$$
leads to contradiction (get $n$ such that $|f(y)-f_n(y)| < \varepsilon/2$ and $|f(y)-f_n(y)| > \varepsilon/2$, and then use the fact that $f$ is uniformly continuous on $[a,b]$).
