Why could it be assumed that the standard enthalpy change is independent of temperature down to 25 °C? For the reversible reaction of pyruvic acid to 2,2-dihydroxypropanoic acid, we calculated enthalpy using NMR and extrapolated from area (integration) and peaks etc. 

We then plotted $\ln K$ vs $1/T.$ I am confused by this questions because I thought enthalpy was temperature dependent. 

I guess my question was unclear. The exact question we are asked is:


  With the assumption that the standard enthalpy change is independent of $T$ down to 25 °C, calculate an extrapolated value for $K$ (25 °C) and compare this to your experimental $K$ value determined at 25 °C. Is this a fair assumption? Why?


I do not understand if this assumption is fair and why?
Generally speaking you can write that 
$$\Delta H^\circ(T) = \Delta H^\circ(T_\mathrm{ref}) + \int_{T_\mathrm{ref}}^T \Delta C_p \,\mathrm dT \tag{1}$$
where $T_\mathrm{ref}$ might be $\pu{25 ^\circ C}$ and the pressure is 1 bar (in your case, approximately 1 bar). 
If your temperature window is not too large you may get away with assuming $\Delta H^\circ$ is constant. This amounts to assuming that the second term in the equation, dependent on $\Delta C_p$, is small compared to the first. If it isn't then rather than using the integrated van't Hoff equation 

$$\ln\left(\frac{K_2}{K_1}\right)=\frac{-\Delta H^\circ}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right) \label{eqn:2}\tag{2}$$
maybe you should attempt a polynomial fit to $\ln K_\mathrm{eq}$ and taking the derivative of the polynomial wrt $T^{-1}$ to evaluate $\Delta H^\circ$ as a function of $T$.

To test whether \eqref{eqn:2} is appropriate you might evaluate the quality of a fit to that equation or check the linearity of a plot of $\ln K$ versus $T^{-1}$.