Using Bernstein polynoms it can be proven that $(X, \|\cdot\|) := (C^0([0,1]), \|\cdot\|_{C^0([0,1])})$ is a seperable vector space.
However, here is my “proof” that this space is not seperable:
Take an arbitrary series $(f_n)_{n \in \mathbb{N}} \subset X$. Let $f$ be a function in $X$ with
$$ f(x) = f_n(x) + 1 \qquad \text{for $x \in \left[\frac{1}{2n+1}, \frac{1}{2n}\right] =: M_n$}. $$
(The existence of such a function can be easily seen.)
Then one has $$ \|f_n - f\|_X \ge \|f_n - f\|_{C^0(M_n)} = 1 \not\rightarrow 0 \qquad \text{for $n \rightarrow \infty$} $$ ie. $\{f_n : n \in \mathbb{N}\}$ is not a dense subset of $X$. As the series was chosen arbitrarily, $X$ is not seperable.
Where is the flaw in this “proof”?