Is there always one integer between these two rational numbers? It appears that for each integer $k\geq2$, there is always one integer $c$ that satisfies the inequalities below. Can this be proved?
$$\frac{3^k-2^k}{2^k-1}<c\leq \frac{3^k-1}{2^k}.$$
Note that for $k\geq2$ the lower bound is always a proper fraction and will never match an integer.  
Edit 28/1/2018 I have a short proof here on Overleaf. <--- It's done! Edit #28 is perfect!
edit It looks like Waring's problem already has a solution. The sequence: https://oeis.org/A060692 "Corresponds to the only solution of the Diophantine equation 3^n = x*2^n + y*1^n with constraint 0 <= y < 2^n." This is equivalent to our $a+c$ for the Waring if statement. Oops, there is no proof that the diophantine equation is always $\leq 2^n.$ Here is a proof sketch of Waring's
Here is a brute force function:
aplusc[k_] := Module[{c}, c = 1; While[0 < 3^k - 2^k (++c)]; 3^k - 2^k (--c) + c]
where we increment $c$ until the calculation becomes negative, then we decrement by one to get $c$. We recalculate using that $c$ and add them together. A060692(k) equals this value.  
Closed form: here. The brute force function illustrates the sawtooth pattern.  
We can also create a(k), b(k), and c(k) using the same module. And we can use b(k) for the proof. Trivially, we can show b$(k) + 1 \leq 2^k.$
edit. Some more information and references: https://oeis.org/A002379
  PM
 A: We look at the small difference $\delta(k)$ of the the two fractions (which become very near by higher $k$)
$$\frac{3^k-1}{2^k-1}-\left(\frac{3}{2}\right)^k=\delta(k). \tag 1$$
Actually $\delta(k)$ is smaller than $1$ and approaches zero with higher $k$ which can be seen when expanded: 
$$\delta(k)=\frac{3^k-2^k}{2^k(2^k-1)}  \lt 1 $$ and is of order$ (3/4)^k$.     
If we find now, that also the fractional values of the terms in eq (1) equal $\delta(k)$ thus  if we have that
$$\left\{ \frac{3^k-1}{2^k-1} \right\}-\left\{ \left(\frac{3}{2}\right)^{k} \right\}= \delta(k), \tag 2$$ 
then it is obvious, that the two terms in (1) have also a common floor and we can write
$$\left\lfloor {3^k-1 \over 2^k-1}\right\rfloor=\left\lfloor\left({3 \over 2}\right)^k\right\rfloor. \tag 3$$

The truth of this equality (3) for all $k \geq 2$ is what we want to show.       


We introduce now shorter notations $$D(k)=\delta(k) \cdot 2^k =\frac{3^k-2^k}{2^k-1}=\frac{3^k-1}{2^k-1} -1  \tag {4.1}$$ and
$$ E(k)=\delta(k) \cdot \frac{(3^k-1)(2^k-1)}{3^k-2^k} = {3^k-1\over 2^k}
\tag {4.2} $$
and conjecture, that there is always an integer $c$ between them:
$$D(k) \lt c \le E(k) \tag 5$$  

If that is indeed the case then we can write
$$\therefore\ c= \left\lfloor\frac{3^k-1}{2^k-1}\right\rfloor=\left\lfloor\left(\frac{3}{2}\right)^k\right\rfloor \text{for }k\geq2.\ \square$$
(Note: we didn't the required proof here, which actually seems out of reach. The motivation of this "answer" was just to make the original OP's claim and its ideas nicer to read)
A: It seems to me, that a sufficient proof for your first inequality is given in Zudilin,W., A  new  lower  bound  for  ||(3/2)^k|| (manuscript at 2005$\,^\dagger$). 
There he refers to a proof, where your $0.5^k$ (extracted from your ${3^k - 1\over 2^k}$ by writing ${3^k \over 2^k}-{1\over 2^k}=1.5^k-0.5^k$) was even replaced by the larger value $0.577^k$ . This means, that the integer value $c = \lfloor 1.5^k \rfloor$ is proven smaller than $ 1.5^k-0.577^k$. (The      value $0.75^k$ in the Waring-conjecture however is still out of reach) 
The left expression in the lhs inequality in your concatenated inequality is larger than  rhs$-1$ (obvious by rewriting ${3^k-2^k\over 2^k-1}={3^k-1\over 2^k-1}-1 $ and then by expanding the geometric series) and the distance between the lhs and the rhs tends with increasing $k$ quickly towards $1$ and thus at most one integer value can be in the interval between lhs and rhs (namely the value $c=\lfloor (3/2)^k \rfloor $).   
So I assume that with some additional work your full (concatenated) inequality might be provable with elementary means.              

$\,^\dagger$ Preprint for "Journal de Theorie des Nombres de Bordeaux", it is available on Zudilin's homepage here.       

A view into S. Finch's book "mathematical constants" gives some quick insight, see this link to google-books:
A: We show that $\left\lfloor\frac{3^n-1}{2^n-1}\right\rfloor$ and $\left\lfloor \left(\frac{3}{2} \right)^n\right\rfloor$ have a common floor when $\frac{3^n-1}{2^n-1}-\left(\frac{3}{2} \right)^n =$ $ \left\{\frac{3^n-1}{2^n-1}\right\}-\left\{\left(\frac{3}{2} \right)^n\right\} = \delta(n),$ where $\left\{\cdot\right\}$ are the fractional parts and $\delta(n)$ is the difference function.
We verify that the order of the fractions, $\frac{3^n-1}{2^n-1}>\left(\frac{3}{2} \right)^n,$ is true for $n\not=0$ and set $n\geqslant1$ to constrain to the positive numbers.
Notice that the denominators are consecutive integers and therefore relatively prime to each other. The lower denominator, $2^n-1,$ must be greater than one, so we set $n\geqslant2$ as our final constraint.
By the ratio test, it is easy to show $\frac{3^n-1}{2^n-1}$ and $\left(\frac{3}{2} \right)^n$ are always proper fractions for $n\geqslant2$ and therefore they always have fractional parts.
We define $\delta(n):=\frac{3^n-2^n}{4^n-2^n}$ and its one's complement, $1-\delta(n),$ as $\frac{4^n-3^n}{4^n-2^n}.$ We define the interval which contains the common floor, $x,$ as lower bound, $\frac{3^n-1}{2^n-1}-1$ and upper bound, $\left(\frac{3}{2} \right)^n.$ From this we can show that the upper bound minus the lower bound is equal to $\frac{4^n-3^n}{4^n-2^n}.$
The RHS below is $\left\{\left(\frac{3}{2} \right)^n\right\}+\left(1- \left\{\frac{3^n-1}{2^n-1}\right\}\right).$  
We assume a common floor: Let $(n|x)\in \mathbb{Z},\ x\geqslant 1,\text{ and } n\geqslant 2,$
$\frac{4^n-3^n}{4^n-2^n}=\frac{3^n-2^n x}{2^n}+ \frac{\left(2^n-1\right) x+2^n-3^n}{2^n-1},\ \ \text{expand; subtract LHS from both sides}\Rightarrow$
$\left(\frac{3}{2}\right)^n-\frac{3^n}{2^n-1}+\frac{1}{2^n-1}+1=\frac{2^n
     x}{2^n-1}-\frac{x}{2^n-1}+\left(\frac{3}{2}\right)^n-\frac{3^n}{2^n-1}+\frac{1}{2^n-1}-x+1;\ \ \Rightarrow$
$0=\frac{2^n
     x}{2^n-1}-\frac{x}{2^n-1}-x,\ \ \text{simplify}\Rightarrow\text{True.}$     
We assume different floors: Let $(n|w|x)\in \mathbb{Z},\ w\geqslant 1,\ x\geqslant 1,\text{ and } n\geqslant 2,$
$\frac{4^n-3^n}{4^n-2^n}=\frac{3^n-2^n x}{2^n}+ \frac{\left(2^n-1\right) w+2^n-3^n}{2^n-1},\ \ \text{expand; subtract LHS from both sides}\Rightarrow$
$\left(\frac{3}{2}\right)^n-\frac{3^n}{2^n-1}+\frac{1}{2^n-1}+1=\frac{2^n
 w}{2^n-1}-\frac{w}{2^n-1}+\left(\frac{3}{2}\right)^n-\frac{3^n}{2^n-1}+\frac{1}{2^n-1}-x+1;\ \ \Rightarrow$
$0=\frac{2^n w}{2^n-1}-\frac{w}{2^n-1}-x,\ \ \text{simplify}\Rightarrow w=x.$
Next, we craft two unit-intervals within which the bounds reside: 
\begin{align}
&\left(\frac{3^n-1}{2^n-1}-1-\left\{\frac{3^n-1}{2^n-1}\right\},\ \frac{3^n-1}{2^n-1}-\left\{\frac{3^n-1}{2^n-1}\right\}\right),\ \ (1)\\ &\left(\left(\frac{3}{2} \right)^n-\left\{\left(\frac{3}{2} \right)^n\right\},\ \left(\frac{3}{2} \right)^n+1-\left\{\left(\frac{3}{2} \right)^n\right\}\right).\ \ (2)
\end{align}
Since lower bound of (1) $<\left(\frac{3}{2} \right)^n-1$, upper bound of (2) $>\frac{3^n-1}{2^n-1},$ and the remaining bounds equal $x,\text{ we have a common floor because }x$ is the greatest integer less than the fraction, which is the floor definition.$\ \ \ \square$
Edit 6/1 I forgot to show that $x$ was the floor.
