Is it true that if $T:X\to X$ is contractive and has a fixed point, then this fixed point is unique (even if $X$ is not complete)?
If $(X,d)$ is a complete metric space and $T:X\to X$ is a contractive maps it follows there is a unique $x\in X$ such that $T(x)=x$.
If $T$ is contractive, and $X$ is no complete then it does not follow that T has a fixed point (consider $T(x)=x^2-2, \; X=\mathbb{Q}$). In other words, does the contractiveness of $T$ cover the uniqueness all by itself?
Yes. If $x_1, x_2 \in X$ were fixed points of $T$ then there is $K<1$ such that \begin{align*} d(x_1,x_2) = d(T(x_1),T(x_2)) \le K d(x_1, x_2) \end{align*} If $x_1, x_2$ were to be different points, then $d(x_1,x_2) \ne 0$ and you would have \begin{align*} d(x_1,x_2) < d(x_1, x_2) \end{align*} A contradiction.