Let L be language of balanced parentheses.
(a) Prove If there are equal number of ('s and )'s and every prefix of w contains at least as many ('s as )'s, then w is in L.
(b) Prove If w is in L, then there are equal number of ('s and )'s and every prefix of w contains at least as many ('s as )'s.
After much thought, I don't seek what I'm supposed to be doing. All I know is that I'm supposed to be using induction.
Here is the grammar that generates L: $S\to SS|(S)|\epsilon$
You are given two different definitions of the language of balanced parentheses:
You need to show that both definitions are the same. The easy direction is showing that each word generated by the grammar satisfies property (1). You do this by induction - either induction on the length or "structural" induction, which in this case is induction on the number of derivation step (base case: $S \to \epsilon$, steps: $S \to SS$, $S \to (S)$). The other direction is more complicated. You need to use complete induction; given a word $w$ satisfying (1), figure out which production needs to be applied first, and which parts of the word it generates, and then appeal to the induction hypothesis.
In fact you are given two languages
You have to prove that these two languages/sets are equal. One way to prove it is to demonstrate that the grammar you indicated in your post does generate both languages.
Another way is to prove $L_1 \subset L_2$ and $L_2 \subset L_1$.
I would go about this as following:
First I show that if a string $s$ consists of balanced parentheses then it is generated by the grammar.
Proof: (by induction on the length of the string)
Base case: the string $()$. It is generated by $S\Rightarrow (S) \Rightarrow ()$.
Induction:
Case 1: $s = \alpha\beta = (...)(...)$. Then by induction on the length of strings we know that $S\Rightarrow^* \alpha = (...)$ and $S\Rightarrow^* \beta = (...)$, so we can generate $s$ by $S \Rightarrow SS \Rightarrow^*(...)(...)$.
Case 2: $s = ((...)) = (\alpha)$. Then by induction we know $S \Rightarrow^* \alpha$, and so we can derive the whole string by $S \Rightarrow (S) \Rightarrow^* ((...))$.
Then I would show that any string generated by the grammar consists of balanced strings.
Proof: by induction on the length of a derivation.
Base case: $n=1$, $S \Rightarrow ()$ is clear. Ignore $S \Rightarrow \epsilon$ since the grammar may be rewritten without $\epsilon$.
Induction: Fix $n$ - length of a derivation leading to terminal strings.
Case 1: Start with $S \Rightarrow SS$. Both S's turn into terminal strings $\alpha$ and $\beta$ respectively in fewer than $n$ steps and so both are strings of balanced parentheses. Hence $\alpha \beta$ is a string of balanced parentheses
Case 2: Start with $S \Rightarrow (S)$. $S$ turns into a terminal string $\alpha$ in fewer than $n$ steps and so is a string of balanced parentheses. Thus $(\alpha)$ is a string of balanced parentheses.
Therefore, the grammar generates only and only ALL strings with balanced parentheses.
Analogously for the language $L_2$.
Hint. Since you don't have a formal definition for $L$, I suggest to use the following one: $L$ is the context-free language generated by the grammar $$ S \to SS + (S) + 1 $$ where $1$ is the empty word. Now you can try to prove (1) and (2). In my opinion, (2) might be easier.
