Isometry on a dense sub-space of a Banach space?

Let $X$ be a Banach space and let $D$ be a dense sub-space of $X$. I don't know if the following fact is true:

Fact: For every (linear) isometry $T\in\operatorname{Iso}(X)$ and for every $\varepsilon > 0$ there is an isometry (linear) $S\in\operatorname{Iso}(D)$ such that: $\|S-T\|<\varepsilon$.

Thank for any hint.

I've been thinking about this all week, and I swear I have a counterexample. Consider $L^1([0,1])$ with the dense subset $C^{\infty}$. The isometry on the main space is multiplying each function by a piecewise function which is 1 on the interval $[0,1/2)$ and $-1$ on the rest of the unit interval. This is an isometry, but it does not fix $C^{\infty}$, because it introduces large gaps.

Now, there are continuous functions that bridge such gaps in short time, but the steeper a function of norm 1 is at $1/2$, the faster its image under a hypothetical isometry of $C^{\infty}$ would have to bridge the gap to stay a fixed distance away. Then take two functions of norm 1, one not so steep, one very steep; the image of their difference is the difference of their images, which would bridge the gap a little to slowly and thus not be an isometry.

I'm sorry this isn't fleshed out more, and I'd be interested in hearing your thoughts.

I really don't understand your counterexample. To be sure that we speak the same language, i will resume what i understand for your counterexample:

For $X$ you take $L^{1}([0,1])$ and for $D$ you take $C^{\infty}([0,1])$. After you consider the following map:

$g(x)=\left\{ \begin{array}{ccccc} 1& if & x\in [0,\frac{1}{2}) \\ -1& if & x \in (\frac{1}{2},1] \\ \end{array}\right.$

Now for $T$ you take

$\begin{array}{lll} T:& L^{1}([0,1]) &\longrightarrow L^{1}([0,1])\\ &f &\longmapsto gf\\ \end{array}$

But i don't understand why you manage to have a contradiction with this. Please feel free to correct me if my interpretation of you counterexample is not true.

Thank one more