This is probably easy, but I can't think of an answer. Assume $X$ is a Banach space and $A$ is a (not assumed closed) subspace of $X$. Let $T:X \to X$ be a bounded linear operator, which is also injective. If $T(A)$ is dense in $X$, does it follow that $A$ is dense in $X$? 



Density of sets whose image is dense
Here's an explicit example for what Bill's answer proposes.


Let $T\_1$ be the operator which maps $e\_n$ to $4^{-n} e\_{n-1}$ for $n \ge 2$ and maps $e\_1$ to $0$. That is, $T\_1 \left(\sum\_{n \ge 1} a\_n e\_n\right) = \sum\_{n \ge 1} a\_{n+1} 4^{-(n+1)} e\_n$. Note $T\_1$ is injective on $A$, and $T\_1 A$ is dense (it contains $c\_{00}$).


Let $y = \sum\_{n} 2^{-n} e\_n$. Note $y \notin T\_1 A$. Define $T$ by $T e\_n = 4^{-n} e\_{n-1}$ for $n \ge 2$ and $T e\_1 = y$. Now we still have that $TA = T\_1 A$ is dense. To see $T$ is injective, suppose $Tx = 0$. Then we can write $x$ uniquely as $x = a\_1 e\_1 + u$ where $u \in A$. Thus $T x = a\_1 y + T\_1 u$. But $T\_1 A$ is a vector space not containing $y$, so if $Tx=0$ we conclude that $a\_1 = 0$ and $T\_1 u = 0$. But $T\_1$ was injective on $A$ so $u = 0$. Thus $x=0$.


