I am new to proofs so feel free to point out any errors or post a better one

For your first proof, not bad! However, while $248$ is correct, you only showed that it suffices; your proof for why it is minimial needs to be attacked from a different angle. The part where you say "Any other number system cannot reach all the numbers up to 1500" needs to be proved, and in fact this is the toughest part of the problem. If you would like to see a nice solution for this, here is the link to do so: http://www.kalva.demon.co.uk/usa/usoln/usol923.html Keep it up!

Denote $S_n$ as $a_1+a_2+\cdots+a_n$. Now for all $S_n<1500$, there must be some set of $a_i$ such that the sum equals $S_n+1$. However, this set of $a_i$ must include $a_k$ with $k\geq n+1$ since $a_1+\cdots+a_n=S_n<S_n+1$. Thus $ a_{n+1}\leq k<S_n+1\implies a_{n+1}\leq S_n+1$. Now obviously $a_1=1$ and $a_2=2$. Lemma: $a_n\leq 2^{n-1}$ We proceed by induction: Base Case: $a_1=1\leq 2^{1-1}=1$. Inductive hypothesis: $a_k\leq 2^{k-1}$. Now since $a_{n+1}\leq S_n+1=a_1+\cdots+a_n+1\leq 1+2+2^2+\cdots +2^{n-1}+1\leq 2^n$ as desired. End Lemma Now $S_11\geq 1500$ otherwise $1500$ cannot be the sum of any set of $a_i$, therefore \[1500\leq S_{11}=S_{10}+a_{11}\leq 2S_{10}+1\implies S_{10}\geq \frac{1499}{2}\implies S_{10}\geq 750\] Thus $a_{10}+a_9+S_8\geq 750$ and since $S_8=a_8+\cdots+a_1\leq 2^7+\cdots+1=2^8-1$, we have \[750\leq a_{10}+a_9+S_8\leq a_{10}+a_9+2^8-1=a_{10}+a_9+255\implies a_{10}+a_9\geq 495\] Also, $a_{10}> a_9\implies 495\leq a_{10}+a_9\leq 2a_{10}\implies a_{10}\geq 247.5$ so $a_{10}\geq 248$ since $a_i$'s are integers. Now I will show that such a sequence is achievable. Let $a_i=2^{i-1}$ for all $i\leq 8$, so all integers from $1$ to $255$ are achievable (binary representation). Now let $a_9=247$ so all numbers from $247$ to $247+255=502$ are achievable by adding on $a_9$. Then let $a_{10}=248$, so all numbers from $247+248=495$ to $247+248+255=750$ are achievable by adding on $a_{10}$. Then let $a_{11}=750$ now all numbers from $750$ to $750+750=1500$ are achievable by adding on $a_{10}$. Therefore $a_{10}=248$ is achievable and thus the lowest possible value of $a_{10}$ is $\boxed{248}$.