Prove that $X(a)$ is closed and boundary set in metric space $(C[0,1], d_{\sup})$ 
A.
  For each $a \in \mathbb R$ let
  $$X(a) = \{ f \in C[0,1] \mid f(0) = a \} $$
  Prove that $X(a)$ is closed and boundary set in metric space $(C[0,1], d_{\sup})$
B.
  Let $$ X = \bigcup_{q \in \mathbb Q} X(q) $$  Prove that subspace $X$ of space $(C[0,1], d_{\sup})$ is not completely metrizable 

My attempt
A. We know that if $(C[0,1], d_{\sup})$ is metric space and $X(a) \subset C[0,1]$ then
$$\overline{X} = \{ g \in C[0,1] : d_{\sup}(g, X(a)) = 0 \} = \{  g \in C[0,1] : \inf \{ d_{\sup}(g, f) : f \in X(a)\} = 0 \ \}$$ 
So $\overline{X(a)}$ is created from that $g \in C[0,1]$ for which exists sequence $g_n$ where $\forall_j g_j \in X(a)$ which is convergent to $f$.
But there I am not sure how to prove that $X(a)= \overline {X(a)}$ and how to deal with boundary proof.
B.
If the thesis is true, we have to prove that metric $d_{\sup}$ doesn't generate space topology $X$ and that this metric space $(X, d_{\sup})$ is not complete 
 A: For A, it is easy to show that $\phi : C[0,1] \to \mathbb{R}$ given by $\phi(f) = f(0)$ is continuous w.r.t. $d_\text{sup}$ so
$$X(a) = \phi^{-1}(\{a\})$$
is closed as a preimage of a closed set via a continuous function.
Notice that $\partial X(a) = X(a)$. Namely, we have
$$\partial X(a) = \overline{X(a)} \cap \overline{X(a)^c} = X(a) \cap C[0,1] = X(a)$$
since $$\overline{X(a)^c} = \overline{\{f \in C[0,1] : f(0) \ne a\}} = C[0,1].$$
Indeed, let $f\in C[0,1]$. If $f(0) \ne a$ then clearly $f \in \overline{X(a)^c}$, and if $f(0) = a$ then $f+\frac1n$ is a sequence in $X(a)^c$ which converges to $f$ uniformly.
We conclude that $X(a)$ is a boundary of a closed set so in particular it is nowhere dense.
For B, we have that $X$ is a union of countably many nowhere dense sets. If $X$ were completely metrizable, this would contradict the Baire Category Theorem.
A: If $f \notin X(a)$ then $f(0) \neq a$. Let $r = |f(0)-a| >0$ and note that
$d_{\text{sup}}(f,g) < r$ implies $|f(0)-g(0)| \le d_{\text{sup}}(f,g)< r = |f(0) -a|$ so that $g(0) \neq a$, hence $g \notin X(a)$ and so $B(f,r) \cap X(a) = \emptyset$. So $X(a)$ is closed.
Also for any $f \in X(a)$ and $r>0$, $g:=f+\frac{r}{2} \in B(f,r)$ and $g(0)=\frac{r}{2} \neq 0$ so $B(f,r)$ intersects $X(a)^\complement$ for any $r>0$ and this implies $X(a)$ is a "boundary set" (i.e. nowhere dense).
b) follows from Baire's theorem, of course, as $\Bbb Q$ is countable.
