I want to show $$ \lim_{x \rightarrow \infty} (x+k)(F(x+c)-F(x)) = 0$$ where $F$ is a cdf of a continuous distribution with finite variance, and $c$ and $k$ are positive constants.
Using the method described here, I can show $ \lim_{x \rightarrow \infty} x(F(x+c)-F(x)) = 0$, and $ \lim_{x \rightarrow \infty} (x+k)(F(x+k+c)-F(x+k)) = 0$, but I can't be certain if my result follows from these. Does it? If not, what are some conditions under which it would?
For a general distribution the method you have described is not applicable since $F$ need not even be continuous.
The result is not true in general. You can show that if $F(x)=1-\frac {\ln \, 2} {\ln (1+x)}$ for $x \geq 1$ and $0$ for $x <1$ the this property is not satisfied.
The result is true if the expectation is finite. This follows from Chebychev's inequálity: $P(X>x) \leq \frac {E|X|I_{\{X>x\}}} x$ for $x >0$.