Problem: Show that $f_n^2$ converges to $f^2$ uniformly whenever $\{f_n\}$ is a sequence of continuous functions on a compact metric space $K$ converging uniformly to a function $f$.
This seems almost too simple so I want to check my proof: We know $$ |f_n^2(x)-f^2(x)|=|f_n(x)-f(x)| \; |f_n(x)+f(x)| $$ As $f_n \to f$ uniformly and the $f_n$ are continuous, we know that $f$ is continuous. Since $f$ is continuous on the compact space $K$, we know that $f$ is bounded, say $|f(x)| \leq B$. We know also that the $f_n$ are continuous on the compact space $K$, so that $|f_n(x)| \leq B_n$ for some $M_n \in \mathbb{R}$. Let $M_n=\max\{B_n,B\}$. As $f_n \to f$ uniformly, given $\epsilon>0$, there is a $N \in \mathbb{N}$ such that $|f_n(x)-f(x)|< \epsilon /(2M_N$). Then $$ |f_n^2(x)-f^2(x)|=|f_n(x)-f(x)| \; |f_n(x)+f(x)| < \epsilon/(2M_N) \cdot 2M_N = \epsilon $$