$f_1,f_2,\ldots$ continuous on $[0,1]$ s.t $f_1 \geq f_2 \geq \ldots$ and $\lim_{n\to\infty}f_n(x)=0$ . 
Let $f_1,f_2,\ldots$ be continuous functions on $[0,1]$ satisfying $f_1 \geq f_2 \geq \cdots$ and such that $\lim_{n\to\infty} f_n(x)=0$ for each $x$. Must the sequence $\left\{f_n\right\}$ converge to $0$ uniformly on $[0,1]$?

I want to say yes and I want to try to invoke Arzela-Ascoli Theorem.  My intuition tells me that for large $n$ we have that the slope of $f_n$ on small enough interval has slope less than 1. So $f_n$ must be Lipschitz and hence equicontinuous. So it must have a uniformly convergent subsequence, but that gives us that $f_n$ must converge uniformly to $0$.
Any help would be appreciated.
 A: Let $C_{n,\varepsilon} = \{x\in[0,1] : f_n(x) \ge \varepsilon\}$.  Then $C_{1,\varepsilon} \supseteq C_{2,\varepsilon}\supseteq\cdots$ and these are compact sets.  Since $\lim\limits_n f_n=0$, their intersection is empty (assume $\varepsilon>0$, of course).  Therefore the intersection of some finite subcollection is empty.  The sequence of $C$s reaches $\varnothing$ after only finitely many steps.  So the sequence of $f$s is uniformly $<\varepsilon$ after finitely many steps.  Hence you have uniform convergence.
A: Define the sequence $a_n=\max\limits_{x \in [0,1]} f_n(x)$. The fact that $f_n$ is decreasing proves that $a_n$ is also decreasing and bounded by zero, thus convergent. Denote by $L=\lim a_n$. If $L=0$ then $f_n$ uniformly converges to zero.
For every $n$ there is an $x_n$ such that $f_n(x_n)=a_n$. Passing eventually to a subsequence we can say that $x_n$ is convergent and denote $x=\lim x_n$. 
Pick $\varepsilon >0$. For $p$ large enough, from continuity of $f_n$ we get that
$$ |f_n(x_{n+p})-f_n(x)|<\varepsilon $$
Therefore, we can write
$$ f_{n+p}(x_{n+p})-f_n(x)\leq f_n(x_{n+p})-f_n(x) <\varepsilon$$
which is
$$ a_{n+p}-f_n(x)<\varepsilon$$
Take $p \to \infty$ and get
$$ L -f_n(x)<\varepsilon$$
Take $n \to \infty$ and get $L<\varepsilon$. Finally for $\varepsilon \to 0$ we get $L=0$.
