Let $f_n\to 0$ be a sequence of functions converging pointwise to $0$. Suppose $|f_n'(x)|\leq 1$ for all $n$ and all $x\in(0,1)$. Show $f_n\to0$ uniformly on $[0,1]$.
My solution: Let $\epsilon>0$. Let $x_0\in(0,1)$. Since $|f_n'(x)|\leq 1$, we have $$ |f_n(x_0)-f_n(x)|<\epsilon/2 $$ whenever $|x_0-x|<\epsilon/2$. Since $f_n(x_0)\to0$, there is an $N_{x_0}$ such that $|f_n(x_0)|<\epsilon/2$ whenever $n\geq N_{x_0}$. Putting this together, we have $|f_n(x)|<\epsilon$ for all $n\geq N_{x_0}$ and $|x-x_0|<\epsilon/2$. This shows that for every point in $(0,1)$, there is a neighborhood around it such that $f_n\to0$ uniformly on that neighborhood. Cover $[0,1]$ by such neighborhoods. Since $[0,1]$ is compact, we have a finite subcover $N_1,\ldots,N_k$. Now for any $\alpha>0$, there is a threshold $J_i$ such that $|f_j(x)|<\alpha$ for $j\geq J_i$ on $N_i$. Take $J=\max(J_1,\ldots,J_k)$. Then for $j\geq J$, we have $|f_j(x)|<\alpha$ for all $x\in[0,1]$. Thus the convergence is uniform. Is this correct?