Unique continuos linear function given a continuous function from a dense space in X to Y (Y is a Banach Space). Let $X$ be a normed space, let $Y$ be a Banach Space, let $D\subseteq X$ be a dense linear subspace of $X$ and let $L:D\rightarrow Y$ be a continuous linear function. Then there is a unique continuous linear function $M:X\rightarrow Y$ such that $M|_D=L$.
The task is to prove this theorem. I have no idea where to even start, any direction you can give me would be greatly appreciated. 
edit: " Banach Space, let $D\in X$ " [;\rightarrow ;] "" Banach Space, let $D\subseteq X$ ""

More general version: $X$ is a metric space, $Y$ a complete metric space,  $D\subseteq X$ is dense, and $L:D\to Y$ is uniformly continuous. A continuous extension $M: X\to Y$ can be defined by 
$$\{M(x)\}=\bigcap_{n=1}^\infty \overline{L(B(x,1/n))} \tag1$$
where $B(x,1/n)$ is open ball centered at $x$ with radius $1/n$. Uniform continuity guarantees that the intersection in (1) has at most one element, while completeness makes sure it's exactly one element. It is straightforward to check that $M$ is uniformly continuous (with the same modulus of continuity). And in the context of linear maps between normed spaces, one can check that $M$ is linear as well.
