Let $f_n, f:[a, b] \rightarrow \mathbb{R} $ for $n=1,2,3,...$ and suppose $\lim_{n\rightarrow \infty} f_n(t) =f(t) $, ie for all $t \in [a,b], \epsilon>0$ there exists $N(t, \epsilon) \in \mathbb{N} $ such that for $n>N, |f_n(t) - f(t) |<\epsilon. $
Consider the set $A = \lbrace N(t, \epsilon) :t \in [a, b] \rbrace$. Then for all epsilon greater than zero why wouldn't it be true for $ n>\sup A$ (which is an integer), that $|f_n(t) - f(t) |< \epsilon$, for all $t\in[a, b] $? (ie the sequence is uniformly convergent)
Consider the sequence $f_n(t)=t^n$ on $[0,1]$ then $f(t)=0$ for $t\in [0,1)$ and $f(1)=1$. Then $$\sup_{t\in[0,1]}|f_n(t)-f(t)|=\sup_{t\in[0,1)}t^n=1$$ and the convergence is not uniform in $[0,1]$.
What is $N(t,\epsilon)$ in this case? What is $\sup \{ N(t, \epsilon) :t \in [a, b] \}$?