Subspaces of separable normed spaces

Let $X$ be a separable normed space. Is it true that every subspace is separable? If it was Hilbert space I would take the dense set and then their projections. It sounds trivial but I cannot prove or disprove it...

Thank you.

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Hint: A metric space is separable if and only if it is second countable. Second countability passes to subspaces. – t.b. Aug 9 '12 at 15:09
I see so it is seperable because every open set of a subspace is of the form $Y \cap V$ where V is an open set of $X$ – clark Aug 9 '12 at 15:14
A metric space is separable IFF it doesn't contain uncountable family of points with pairwise distance greater than some $\delta$. Corollary - subspace of separable space is separable. – Norbert Aug 9 '12 at 15:51
My problem was how to pass from the topology of the subspace to the whole topology space – clark Aug 9 '12 at 15:53

Let $X$ be a separable metric space and $x_n$ a dense subset. Let $a_m$ be an enumeration of the positive rational numbers and let $V_{(n,m)}=\{y \in X \mid d(y,x_n) < a_m \}$. This is a countable base of X.
Indeed, take an open set $U$ of $X$. For $y_0 \in U$ there is $\epsilon >0$ such that $B(y_0, \epsilon) \subset U$. Choose $x_{n_0}$ such that $d(x_{n_0},y)< \frac{ \epsilon }{4}$ and $a_{m_0}$ such that $\frac{ \epsilon }{4}< a_{m_0} < \frac{ \epsilon }{2}$.Then $y \in V_y=V_{ (n_0 , m_0)} \subset U$ and $$U= \bigcup_{y \in U}V_y.$$
Small nitpick: better take all rational radii (instead of only those of the form $1/m$) because there need not be a number of the form $1/m_0$ between $\epsilon/4$ and $\epsilon/2$. – t.b. Aug 9 '12 at 15:52