We want to show that, letting $(f_n)_{n\in\mathbb{N}}$ be a sequence of functions on $[0,1]$ that's uniformly convergent, then there exists $M>0$ such that $\forall n\in\mathbb{N}$, $|f_n(x)|\leq M$. I've seen some proofs of this here but I'm a bit confused on my own solution, which does not use the fact that continuous functions that uniformly converge must converge to a continuous function since it hasn't been introduced in my text at that point.
My proof attempt:
Let $(f_n)_{n\in\mathbb{N}}$ be a sequence of continuous functions on $[0,1]$ that's uniformly convergent to some continuous function $f(x)$ and let $\epsilon=1$. Then, we know that, by Cauchy property of uniform convergence, $\exists N$ such that $\forall n\geq N$, $$\big||f_n(x)|-|f_N(x)|\big|\leq |f_n(x)-f_N(x)|<1$$ (where the first part being reverse triangle inequality). Then $\forall n\geq N$, we have $$-1<|f_n(x)|-|f_N(x)|<1$$ and hence $$|f_n(x)|<1+|f_N(x)|$$ Further, by the fact that $[0,1]$ is compact, $f_N$ is continuous (since the sequence of $f_n$ is continuous), and Extreme Value Theorem, $|f_N(x)|$ is bounded. Same logic applies for other $f_n(x)$ which are continuous and thus bounded on $[0,1]$. We then let: $$|f_N(x)|=M_N$$ and for each $i<N$, let $M_i$ be such that: $$|f_i(x)|\leq M_i\ (\forall x\in[0,1])$$ Taking $M=\max\{M_1,\ldots, M_{N-1}, M_N+1\}$, we have $|f_n(x)|\leq M$ for all $n\in\mathbb{N}$.
Was simply wondering if this argument is valid as it doesn't take some $f$ where the $f_n$ are converge uniformly towards. Thanks!