A question about a closed set Let $X = C([0; 1])$. For all $f, g \in X$, we define the metric $d$ by
$d(f; g) = \sup_x |f(x) - g(x)|$. Show that $S := \{ f\in X : f(0) = 0 \}$ is closed in $(X; d)$.
I am trying to show that $X \setminus S$ is open but I don't know where to start showing that.
I wanna add something more, I have not much knowledge about analysis and I am just self taught of it, what I have learnt so far is just some basic topology and open/closed sets.

I'd try to show that $S$ contains all its limit points. To this end let $f$ be a limit point of $S$. Let $f_n$ be a sequence in $S$ converging to $f$ in the sup norm. 
Now we show that $f$ is also in $S$:
By assumption, for $\varepsilon > 0$ you have that $\sup_{z \in [0,1]}|f_n(z) - f(z)| < \varepsilon$ for $n$ large enough. In particular, $|f_n(0) - f(0)| = | f(0)| < \varepsilon$ for $n$ large enough. Now let $\varepsilon \to 0$.
