Proof involving uniform convergence of a sequence and functions. Suppose $f_n\to f$ uniformly on $[0,1]$ and $g$ is continuous on $[0,1]$. Prove that the sequence of functions $f_ng$ converges uniformly to the product $fg$ on $[0,1]$.
My draft of a proof is the folloing:
Since $g$ is continuous on $[0,1]$, it is bounded by the Extreme Value Theorem. Since $f$ is bounded and $\{f_n\}$ is uniformly bounded, there is an $M>0$ such that $\text{max}\{|f_n(x)-f(x)|:x\in[0,1],n\in\mathbb{N}\}\leq M$. Given $\epsilon>0$ choose $\delta>0$ so small that $0<x<0+\delta$ or $1>x>1-\delta$ inplies $|g(x)|<\frac{\epsilon}{M}$. By hypothesis, $f_n\to f$ uniformly on $[0+\delta,1-\delta]$. Thus choose $N$ so large that $x\in[0+\delta,1-\delta]$ and $n\geq N$ imply $|f_n(x)-f(x)|<\frac{\epsilon}{C}$. If $n\geq N$ and $x\in[0,1]$, then $$|f_n(x)g(x)-f(x)g(x)|=|f_n-f(x)||g(x)|<\begin{cases}
(\frac{\epsilon}{C})\cdot C=\epsilon \hspace{.5cm} \text{for} \hspace{.5cm} x\in[0+\delta,1-\delta]\\
(\frac{\epsilon}{M})\cdot M=\epsilon \hspace{.5cm} \text{for} \hspace{.5cm} x\not\in[0+\delta,1-\delta].\end{cases}$$
Therefore, $f_ng\to fg$ uniformly on $[0,1]$
Is there anywhere I can improve or are incorrect?
 A: The property holds as soon as $g$ is bounded in $[0,1]$ (and this is true when $g$ is continuous):
$$\sup_{x\in[0,1]}|f_n(x)g(x)-f(x)g(x)|\leq\sup_{x\in[0,1]}|f_n(x)-f(x)|\cdot \sup_{x\in[0,1]}|g(x)|.$$
Therefore, if $f_n\to f$ uniformly in $[0,1]$ then $\sup_{x\in[0,1]}|f_n(x)-f(x)|\to 0$ and, by the above inequality, $\sup_{x\in[0,1]}|f_n(x)g(x)-f(x)g(x)|\to 0$ and we may conclude that $f_ng\to fg$ uniformly in $[0,1]$.
A: Thanks to Robert and B. Mehta I think I am confident in my proof.
Since $g$ is continuous on $[0,1]$, it is bounded by the Extreme Value Theorem. 
In other words, there is a $C>0$ such that $|g(x)|\leq C$ for all $x\in[0,1]$. 
Additionally, there is an $M>0$ such that $\sup_{x\in[0,1]}\{|f_n(x)-f(x)|,n\in\mathbb{N}\}\leq M$. 
Given $\epsilon>0$ choose $\delta>0$ so small that $0<x<0+\delta$ or $1>x>1-\delta$ implies $|g(x)|<\frac{\epsilon}{M}$. 
By hypothesis, $f_n\to f$ uniformly on $[0+\delta,1-\delta]$. 
Thus choose $N$ so large that $x\in[0+\delta,1-\delta]$ and $n\geq N$ imply $|f_n(x)-f(x)|<\frac{\epsilon}{C}$. 
If $n\geq N$ and $x\in[0,1]$, then $$\sup|f_n(x)g(x)-f(x)g(x)|\leq\sup|f_n-f(x)|\cdot\sup|g(x)|\leq\sup|g(x)|\frac{\epsilon}{C}<\epsilon\hspace{.2cm}\text{for}\hspace{.2cm}x\in[0+\delta,1-\delta]$$
Therefore, $f_ng\to fg$ uniformly on $[0,1].$
