Injective bounded linear operator on Banach space maps non-dense set to dense set. Let $A\subseteq X$ be a subspace ($X$ is a Banach space). Let $Y$ be another Banach space. Consider a continuous injective linear functional $f:X\rightarrow Y$. Suppose $f(X)\subseteq \overline{f(A)}$, is it true $X\subseteq \overline{A}$? I kind of feel this is obvious but cannot figure out a proof. (I guess this might be related to Baire Category theorem, but it seems open mapping theorem cannot be applied here.)
The context of the problem is from the proof of Jacod Martingale representation theorem. $\mathcal{S}(A)\subseteq \mathcal{H}^2$ be the closed linear stable space generated by $A$. The injection from $\mathcal{H}^2$ to $\mathcal{H}^1$ satisfying $\overline{i(S(A))}=\mathcal{H}^1$. With the statement above, we can conclude $\mathcal{H}^2=\mathcal{S}(A)$.
Any comments or idea?
Thanks in advance!

This is not true. Let $X=C([0, 1]), Y=L^1([0, 1])$ and $f$ sends a function to itself, which is obviously injective.
As $\|x\|_1=\int_0^1 |x(t)|dt\le\|x\|_{\infty}$, $f$ is clearly bounded.
However $A=\{x\in C([0, 1]): x(0)=0\}$ is closed in $X$, but dense in $Y$.
