Our teacher gave this problem in class and said that the hint to solve this was by using triangle inequality.
Let $\{f_n\}$ be a sequence of continuous real-valued functions on $[0,1]$ converging uniformly on $[0.1]$ to a function $f$. Suppose for all $n \in \mathbb{N}$, there exists $x_n \in [0,1]$ such that $f_n(x_n) = 0$. Show that there exists $x \in [0,1]$ such that $f(x) = 0$.
But I feel, we don't really need the triangle inequality here.
By definition, for every $\epsilon >0,$ $ \exists$ $ N \in \mathbb{N}$ such that $n \geq N$ implies $|f_n(x) - f(x)| \leq \epsilon $ $\forall$ $x \in [0,1]$.
Specifically, for every $\epsilon > 0$, $\exists$ $N \in \mathbb{N}$ such that $n \geq N$ implies $|f_n(x_n) - f(x_n)| \leq \epsilon$.
So, by given information $|f(x_n)| < \epsilon$. And since $f([0,1])$ must be closed, there must exist $x \in [0,1]$ such that $f(x) = 0$.
Can anyone please let me know if my proof is correct ?
If not, where is the fallacy ?