I'll assume you want your $a_i$ to be positive integers, and that you want the statement true for all $k$.
Let $a_1 < a_2 < ... < a_n ...$ be such a sequence. Then for any $n$, and $k\leq n$, the $k$-subsets of $\{a_1,...,a_n\}$ must be distinct. But the sum of any $k$-subset of this set is at most $ka_n$. So there must be at least $n\choose k$ distinct numbers from $1$ to $ka_n$(*), and hence ${n\choose k} \leq ka_n$, or
$$a_n \geq \frac{1}{k} {n\choose k}$$
Now, if $n=2m$ and $k=m$ then you get:
$$a_{2m} \geq \frac{1}{m} {2m\choose m}$$
The right hand side is approximately $\frac{(m+1)2^{2m}}{m^{5/2}\sqrt{\pi}}$. This can be seen via Catalan numbers, see: http://en.wikipedia.org/wiki/Catalan_number .
That would seem to imply that the Fibonacci sequence does not satisfy your original condition. The Fibonacci sequence grows like $\frac{1}{\sqrt 5}\phi^n$, where $\phi = \frac{1+\sqrt{5}}{2} < 2$.
Indeed, if $26 = 21 + 3 + 2 = 13 + 8 + 5$, so this is not true for the Fibonacci numbers when $k=3$.
(*) Edit: Of course, the range is even stronger. $k$ distinct integers from $1$ to $a_n$ add up to a value between $\frac{k(k+1)}{2}$ and $ka_n-\frac{k(k-1)}{2}=\frac{k(2a_n-k+1)}{2}$, for a total of $1+\frac{k(2a_n-2k)}{2}$ distinct possible values. So we can slightly improve the above:
$$\binom{n}{k}\leq 1+k(a_n-k)\\
\frac{1}{k}\binom{n}{k}-\frac{1}{k} +k \leq a_n$$
You want $a+b+c+d = 12$ where $2 \leq a \leq b \leq c \leq d \leq 5$. Let $a_1 = a-2$, $b_1 = b-2$, $c_1 = c-2$ and $d_1 = d-2$. This gives us that $$a_1 + b_1 + c_1 + d_1 = 4$$ where $0 \leq a_1 \leq b_1 \leq c_1 \leq d_1 \leq 3$.
Let $b_1 = a_1 + b_2$, $c_1 = b_1 + c_2$ and $d_1 = c_1 + d_2$. Then we need $$a_1 + (a_1 + b_2) + (a_1 + b_2 + c_2) + (a_1 + b_2 + c_2 + d_2) = 4$$
i.e. $$4a_1 + 3b_2 + 2 c_2 + d_2 = 4$$
where $0 \leq a_1,b_2,c_2,d_2$.
Note that $a = a_1 +2$, $b = a_1 + b_2 + 2$, $c = a_1 + b_2 + c_2 + 2$ and $d = a_1 + b_2 + c_2 + d_2 + 2$
Now all we want is $$4a_1 + 3b_2 + 2 c_2 + d_2 = 4$$
such that $0 \leq a_1 \leq b_1 \leq c_1 \leq d_1 \leq 3$.
This means $a_1 \leq 1$.
If $a_1 = 1$, then $b_2 = c_2 = d_2 = 0$. Hence the solution is $$(a,b,c,d) = (3,3,3,3)$$
If $a_1 = 0$, then $$3b_2 + 2 c_2 + d_2 = 4$$ where $0 \leq b_2,c_2,d_2$.
This means $b_2 \leq 1$.
If $b_2 = 1$, then $c_2 = 0$ and $d_2 = 1$. Hence, the solution is $$(a,b,c,d) = (2,3,3,4)$$
If $b_2 = 0$, then $$2c_2 + d_2 = 4$$ where $0 \leq c_2,d_2$. This gives us $(c_2,d_2) = (2,0)$, $(c_2,d_2) = (1,2)$ and $(c_2,d_2) = (0,4)$. But $d_1 \leq 3$. Hence, the last solution is not possible.
Hence, these now give the solutions
$$(a,b,c,d) = (2,2,4,4)$$
$$(a,b,c,d) = (2,2,3,5)$$
Hence, the only four possible solutions for $a+b+c+d = 12$, with the constraint that $a,b,c,d \in \{2,3,4,5\}$ are
$$(a,b,c,d) = (3,3,3,3)$$
$$(a,b,c,d) = (2,3,3,4)$$
$$(a,b,c,d) = (2,2,4,4)$$
$$(a,b,c,d) = (2,2,3,5)$$
Best Answer
They are called sets with distinct subset sums.