@MartinSleziak So you mean that it is as follows?

Since $X \setminus{ \overline{A}}$ is open, there is an $\epsilon>0$ such that $B_{\rho}(x, \epsilon) \subset X \setminus{ \overline{A}} \Rightarrow \{ y \in X: \rho(x,y)< \epsilon \} \subset X \setminus{\overline{A}}$.

If $a \in A$ then $a \notin X \setminus{\overline{A}}$, so $a \notin \{ y \in X: \rho(x,y)< \epsilon \} \Rightarrow \rho(x,a) \geq \epsilon \Rightarrow \inf \{ \rho(x,a) \geq 0: a \in A\} \Rightarrow d(x,A) \geq 0$

Let T be a linear transformation what's the connection between T(v) and (u,Tv)
Let's say if (u,Tv)=0 can i say T(v)=0?

@MartinSleziak Why do we know the following?

then you know that for each $\varepsilon>0$ there is an $a\in A$ such that $d(x,a)<\varepsilon$.

We have that $S_{||\cdot||_2}:= \{ x \in \mathbb{R}^n: ||x||_2=1\}$.

How can we justify that it is bounded.

Do we just say that if $x \in S_{||\cdot||_2}$ then $||x||_2=1 \leq 1$ and so the set is bounded. How could we justify it more formally?

Suppose that $A $ is symmetric.

We have $g(x)=\langle x, Ax \rangle= \sum_{i=1}^n x_i \sum_{j=1}^n a_{ij} x_j$.

How do we get that $\nabla{g(x)}=2Ax$ ?

I tried it for $n=2$ and I got $\nabla{g(x)}=(2a_{11} x_1+ x_2 (a_{12}+a_{21}), x_1(a_{12}+a_{21})+2x_2 a_{22})$