That's a pretty good proof. The core idea is that you can reduce an instance of the problem (a sequence of gas and cost values) to a smaller instance: if there is enough gas in city i to go from there to (i+1), then you can "merge" city i into city (i+1) by adding their gas and subtracting the cost of going from i to (i+1) from their total gas.
To refine this into a more detailed proof, we can make more formal (1) the notions of inputs and outputs (= solutions) of the problem, and (2) the structure of a proof by induction.
It is quite long winded. Once you are confident that you can spell out proofs to this level of detail, it becomes normal to give high-level outlines instead, which is basically what you did.
Definitions
An input is a triple $(n, \mathrm{gas}, \mathrm{cost})$ where $n \in \mathbb{N}^+$ is the number of cities ($n \ge 1$), and for all $i$ such that $1 \le i \le n$, $\mathrm{gas}_i \in \mathbb{R}^+$, $\mathrm{cost}_i \in \mathbb{R}^+$. Furthermore, valid inputs are those that satisfy: $$\sum_{i=1}^n \mathrm{gas}_i \ge \sum_{i=1}^n \mathrm{cost}_i$$
A solution to a valid input $(n,\mathrm{gas},\mathrm{cost})$ is an index $j_0$ such that, for all $k$ such that $0 \le k \le n-1$,
$$\sum_{i=j_0}^{j_0+k} \mathrm{gas}_i \ge \sum_{i=j_0}^{j_0+k} \mathrm{cost}_i$$
Where, for brevity, the indexing of cities wraps around, if $i > n$, city $i$ = city $i - n$.
We want to prove:
Theorem. Every valid input has a solution.
Proof
For that, we will prove a key lemma:
Lemma. For every valid input $(n,\mathrm{gas},\mathrm{cost})$ where $n > 1$, there exists a valid input $(n-1,\mathrm{gas}', \mathrm{cost}')$ such that if it has a solution, then $(n,\mathrm{gas},\mathrm{cost})$ has a solution.
Postponing the proof of that lemma to the end of this answer, let us first show how the rest of the proof goes using that lemma.
By induction on $n$, we prove that any valid input with $n$ cities has a solution.
Base case: a valid input with 1 city has a solution. Proof: because the input is valid, $\mathrm{gas}_1 \ge \mathrm{cost}_1$, so we have enough gas to take the trip from city $1$ to city $1$, which costs $\mathrm{cost}_1$.
Induction step: We assume that any valid input with $n-1$ cities has a solution (induction hypothesis). We prove that any valid input with $n$ cities has a solution.
- Let $(n,\mathrm{gas},\mathrm{cost})$ be a valid input.
- By the earlier Lemma, there exists a valid input $(n-1,\mathrm{gas}',\mathrm{cost}')$ such that: if $(n-1,\mathrm{gas}',\mathrm{cost}')$ has a solution, then $(n,\mathrm{gas},\mathrm{cost})$ has a valid solution (call this the implication).
- By induction hypothesis, $(n-1,\mathrm{gas}',\mathrm{cost}')$ has a valid solution.
- By the implication which we just got from our application of the Lemma, $(n,\mathrm{gas},\mathrm{cost})$ has a valid solution.
That concludes the proof by induction. To sum up, any valid input has a solution. QED.
Proof of the Lemma
Let $(n,\mathrm{gas},\mathrm{cost})$ be a valid input with $n > 1$. The goal is to (1) construct an input with $n-1$ cities, and (2) map a solution of the smaller input to a solution of the original input.
By definition:
$$\sum_{i=1}^n \mathrm{gas}_i \ge \sum_{i=1}^n \mathrm{cost}_i$$
Construction of a smaller input
There must exist $i_0$ such that $\mathrm{gas}_{i_0} \ge \mathrm{cost}_{i_0}$. (Proof by contradiction: if for all $i$, $\mathrm{gas}_i < \mathrm{cost}_i$, then $\sum_{i=1}^n \mathrm{gas}_i < \sum_{i=1}^n \mathrm{cost}_i$, which contradicts the inequality above.)
We can now define an input with $n-1$ cities, by defining its $\mathrm{gas}'$ and $\mathrm{cost}'$. Intuitively, we merge cities $i_0$ and $i_0+1$, the merged city will be at index $i_0$, and the cities at greater indices will be shifted down accordingly. There is a technicality if $i_0 = n$ which we circumvent by symmetry: we can rotate the indexing of cities beforehand so we can assume $i_0 \not= n$ without loss of generality.
$$\mathrm{gas}'_i = \left\{\begin{array}{ll}\mathrm{gas}_i & \text{if }i < i_0 \\ \mathrm{gas}_{i_0} + \mathrm{gas}_{i_0+1} - \mathrm{cost}_{i_0} & \text{if } i = i_0 \\ \mathrm{gas}_{i+1} & \text{if } i_0 < i \le n - 1 \end{array}\right.$$
$$\mathrm{cost}'_i = \left\{\begin{array}{ll}\mathrm{cost}_i & \text{if }i < i_0 \\ \mathrm{cost}_{i+1} & \text{if } i \ge i_0 \end{array}\right.$$
As a sanity check against off-by-one mistakes, note that for every $i$ such that $1 \le i \le n$, the values $\mathrm{gas}_i$ and $\mathrm{cost}_i$ are each used exactly once in the right-hand sides of the above formulas. In some sense, there is a "conservation law" going on between the bigger input and the smaller one we just constructed and which enables the rest of this proof.
Transforming a small solution to a bigger solution
Next we show that if $(n-1, \mathrm{gas}', \mathrm{cost}')$ has a solution $j_0$, then $(n,\mathrm{gas},\mathrm{cost})$ has a solution. In fact, it will be the same solution $j_0$.
Let $j_0$ be a solution of $(n-1, \mathrm{gas}', \mathrm{cost}')$. For all $k$ such that $0 \le k \le n - 2$,
$$\sum_{i=j_0}^{j_0+k} \mathrm{gas}'_i \ge \sum_{i=j_0}^{j_0+k} \mathrm{cost}'_i$$
The intuition is to split the sum at index $i=i_0$, where the split happens in the definition of $\mathrm{gas}'$. But there are technicalities to be mindful of, the actual index for the split could be $i=n+i_0$, that is the case if $i_0 < j_0$, and there is another "discontinuity" around $i = n-1$, where the indexing of $\mathrm{gas}$ and $\mathrm{gas}'$ wraps around.
By symmetry, we rotate the cities around so that we can assume $j_0 = 0$ without loss of generality. That way there is no "wrap around" to worry about. The above inequality becomes, for all $k$ such that $0 \le k \le n - 2$:
$$\sum_{i=0}^{k} \mathrm{gas}'_i \ge \sum_{i=0}^{k} \mathrm{cost}'_i$$
We want to show that, for all $k$ such that $0 \le k \le n - 1$,
$$\sum_{i=0}^{k} \mathrm{gas}_i \ge \sum_{i=0}^{k} \mathrm{cost}_i$$
There are three cases:
- If $k < i_0$, then those inequalities are identical.
- If $k = i_0$, we apply the former inequality at $k=i_0-1$, and add $\mathrm{gas}_{i_0} \ge \mathrm{cost}_{i_0}$ to it.
- If $i_0 < k \le n - 1$, the following inequalities are equivalent (with the main technicality being the shifting of indices in the first step, when the sums of $\mathrm{gas}'_i$ and $\mathrm{cost}'_i$ become sums of $\mathrm{gas}_i$ and $\mathrm{cost}_i$):
$$\begin{aligned} \sum_{i=0}^{k-1} \mathrm{gas}'_i & \ge \sum_{i=0}^{k-1} \mathrm{cost}'_i \\ \sum_{i=0}^{i_0-1} \mathrm{gas}_i + \mathrm{gas}'_{i_0} + \sum_{i=i_0+2}^{k} \mathrm{gas}_i &\ge \sum_{i=0}^{i_0-1} \mathrm{cost}_i + \sum_{i=i_0+1}^{k} \mathrm{cost}_i \\ \sum_{i=0}^{i_0-1} \mathrm{gas}_i + \mathrm{gas}_{i_0} + \mathrm{gas}_{i_0+1} -\mathrm{cost}_{i_0} + \sum_{i=i_0+2}^{k} \mathrm{gas}_i &\ge \sum_{i=0}^{i_0-1} \mathrm{cost}_i + \sum_{i=i_0+1}^{k} \mathrm{cost}_i \\ \sum_{i=0}^{i_0-1} \mathrm{gas}_i + \mathrm{gas}_{i_0} + \mathrm{gas}_{i_0+1} + \sum_{i=i_0+2}^{k} \mathrm{gas}_i &\ge \sum_{i=0}^{i_0-1} \mathrm{cost}_i + \mathrm{cost}_{i_0} + \sum_{i=i_0+1}^{k} \mathrm{cost}_i \\ \sum_{i=0}^{k} \mathrm{gas}_i & \ge \sum_{i=0}^{k} \mathrm{cost}_i \end{aligned}$$
