Dense subspace of linear space and functional equal to $0$ Let $X$ be a normed space over field $\mathbb{K}$ and let $X_0$ be a linear subspace of $X$. I have to prove that:
$X=\overline X_0 \iff$ For every linear continuous functional $\phi : X\rightarrow \mathbb{K}$ we have $\phi |_{X_0} \equiv0 \implies \phi \equiv 0$ 
My attempt
"$\implies$"
Assume, that there exists $x\in X\setminus X_0$ such that $\phi(x) \ne 0$
$X_0$ is dense in $X$ so for any $\epsilon>0$ I can find $y\in X_0$ such that $\|x-y\|<\epsilon$. So I can find $y\in X_0$ arbitrarly close to $x$ and $\phi(y)=0$. But $\phi$ is continuous. Contradiction.
I have no idea about the second part...
 A: Suppose that there is some non-empty open set $U$ such that $U \cap \overline{X_0} = \emptyset$. We can take $U$ to be of the form
$U=B(\hat{x},\epsilon)$ for some $\epsilon >0$. Since $X_0 $ is a subspace, we have $\hat{x} \ne 0$.
The Hahn Banach theorem gives the existence of some continuous functional $\phi$ and some $\alpha$ such that $\operatorname{re}\phi(u) < \alpha \le \operatorname{re} \phi(y)$ for all $u \in U, y \in \overline{X_0}$.
Use the fact that $X_0$ is a subspace and the form of $U$ to conclude that $\phi(y) = 0$ for all $y \in X_0$, but $\phi(\hat{x}) < 0$.
Addendum: Proof using the version of Hahn Banach in the comments below.
Let $\hat{x}$ be as above and
let $W = \{ \lambda \hat{x} + y | \lambda \in \mathbb{K}, y \in \overline{X_0} \}$. Note that any $w \in W$ has a unique representation $w = \lambda \hat{x} + y$.
Define $f: W \to \mathbb{K}$ by $f(\lambda \hat{x} + y ) = \lambda$.
Note that $\ker f = \overline{X_0}$ is closed (in $W$), hence $f$ is continuous (as a map $W \to \mathbb{K}$), and $f(\hat{x}) = 1$.
The theorem referred to in the comments lets us extend $f$ to $\tilde{f}: X \to \mathbb{K}$ without increasing the norm (in particular, $\tilde{f} \in X^*$, which is what we care about here).
Hence we have a functional $\tilde{f}$ which is zero on $\overline{X_0}$, but non-zero on $X$.
A: The reverse part follows from a corollary of Hahn Banach Theorem:
Assuming the second part, we need to prove $\bar{Y}$=X.
Suppose on contrary:
$\bar{Y}$ $\subset$ X. 
Then there exist a  $x_0$ $\in$ X $\backslash \bar{Y}$.
Observe that $\bar{Y}$ is a closed subspace of X. Then by HBT, there exists f $\in X^*$ such that f($x_0$)=d($x_0$,$\bar{Y}$) and f(y)=0, $\forall$ y $\in$ $\bar{Y}$. Now that we have f $\equiv$ 0 on X $\implies$ d($x_0$,$\bar{Y}$)=0, which clearly implies $x_0$ $\in$ $\bar{Y}$. A contradiction!
