Amplitude not attenuated after bandpass filtering?
You seem to be aware of the Fourier Transform pair that relates a sinc function in the time domain to a rectangle function in the frequency domain, but you don't seem to be able to correctly apply this to your situation. I am going to take one more stab at answering you. Let me know if you have questions.

A sinc filter can be designed using the relationship
\begin{equation}
\DeclareMathOperator{sin}{sin}
\DeclareMathOperator{sinc}{sinc}
\DeclareMathOperator{rect}{rect}
2 B \sinc\left(2 B t\right) \iff \rect\left(\frac{f}{2 B}\right)
\end{equation}
where $\sinc(x) = \frac{\sin(\pi x)}{\pi x}$. This shows us that the desired low pass bandwidth $B$ must be used in our expression to generate the sinc samples. Note that a modulated sinc has a bandwidth equal to $2B$ whereas a sinc centered at DC has a bandwidth of $B$.

In your example, you define several quantities. I am going to give these quantities symbols so I can do some math on them.


  Fcentre = 137.65 MHz


I'll call this $f_c$.


  Fsample = 2 MHz


I'll call this $f_s$.


  Lowcut = 137.65 MHz


I'll call this $f_l$.


  Hightcut = 138.15 MHz


I'll call this $f_h$.

For you to generate the sinc with the correct bandwidth, you would sample the sinc function above with
\begin{equation}
B = \frac{f_h - f_l}{2}.
\end{equation}

This means that your bandpass filter is given as
\begin{equation}
h[n] = \left(f_h - f_l\right) \sinc\left(\frac{f_h - f_l}{f_s} n\right) \exp\left(j \pi \frac{f_h + f_l - 2 f_c}{f_s} n \right)
\end{equation}
where I have ignored the required time shift to make the filter causal (implementable). The amount you shift it by will be determined by the length of the filter. The length of the filter will be determined by how much stopband attenuation you want. 

Note that no finite length filter will yield zero gain in the stopband and all filters have a transition band (a band where the amplitude gradually goes from the desired passband value to the designed stopband value). You can't avoid this.

As a side note, you could have just as easily used the Remez Exchange algorithm (a.k.a. Parks-McClellan) to obtain an optimal filter. The window method is computationally simple, but it is in no sense optimal.