If $D$ is a dense linear subspace of $X$ then $D\to Y$ extends to $X\to Y$ uniquely I am trying to prove the following, but I am not confident in my work.

Let $D$ be a linear subspace of a normed space $X$ that is dense
  in $X$. Let $Y$ be a Banach space. Show that any bounded linear operator $T : D \to Y$ has a unique extension to a bounded linear operator $X \to  Y.$

What I tried. I can prove the lemma
Lemma: Assume that $D$ is a dense subset of $X_1$, that $(X_2, d_2)$ is complete and that $f : D → M_2$ is a uniformly continuous mapping.
Then there exists a unique continuous mapping $F : M_1 → M_2$ such that $F|_D = f$.
Bounded linear operators are uniformly continuous the existence
and uniqueness of a continuous extension follows from the Lemma. If I am correct, I can use the proof of the lemma to construct the desired extension.
 A: What you said is correct. The following proof is essentially the same proof as you will give for your lemma
HINT: For any $x\in X$ let $(x_n)$ be a sequence in $D$ such that $x_n\longrightarrow x$. Since $T$ is linear and bounded, we have
\begin{equation}
\|Tx_n-Tx_m\| = \|T(x_n-x_m)\|\leq\|T\|\|x_n-x_m\|.
\end{equation}
This shows that $(Tx_n)$ is a Cauchy sequence because $(x_n)$ converges. By assumption that $Y$ is a Banach space $(Tx_n)$ converges, say, 
\begin{equation}
Tx_n\longrightarrow y\in Y.
\end{equation}
Define $\tilde{T}$ by 
\begin{equation}
\tilde{T}x=y.
\end{equation}
Now check the following:
(1) $\tilde{T}$ is well defined.
(2) $\tilde{T}$ is linear and is an extension of $T$.
(3) $\tilde{T}$ is bounded. To see this notice that $\|Tx_n\|\leq \|T\|\|x_n\|$ and $x\mapsto \|x\|$ is a continuous mapping.
NOTE: By (2) you will get that $\|\tilde{T}\| \leq \|T\|$ and obviously $\|\tilde{T}\|\geq\|T\|$ because the norm, being defined by supremum, cannot decrease in an extension. So, in fact you have a norm preserving extension, i.e., $\|\tilde{T}\| = \|T\|$.
