I would write out a solution but Wikipedia already has a really nice write-up of the same proof: https://en.m.wikipedia.org/wiki/Catalan_number#Third_proof
Here, they use paths instead of parentheses, but it's very easy to transition between the two. Whenever you move right one step, that's like an open parenthesis, whenever you move up, that's a close parenthesis.
In this way, avoiding violations means at any point in the parenthesis sequence you never have more close parentheses than open, and so you never have more up moves than right moves. It is now easy to see how that is analogous to the path never crossing the middle diagonal.
As for why the exceedance goes down by one, the transform described is analogous to basically taking an open parenthesis that came too late and moving it to the front, thus resolving one of the pairs of parentheses that posed a problem.
Associate to a sequence of parantheses a sequence of $1$, $-1$, by $(\mapsto 1$, $)\mapsto -1$.
Balanced sequence $a_1 a_2 \ldots a_{2n}$ means
$ \sum_1^k a_i\ge 0$ for all $1\le k \le 2n$
$ \sum_1^{2n} a_i = 0$
Let's show first: a sequence of $1$, $-1$, is legal if and only if it is balanced.
Now, legal sequences are sequences obtained as follows: start from the empty sequence, perform several steps of form
$$ A,B \mapsto A, 1,-1, B$$
that is, we insert into the word $A,B$ at some position the piece $1, -1$ ($A$, $B$ are words in $1$, $-1$, possibly empty).
We can see that if $AB$ is balanced, so is the new sequence. Hence, by induction, all legal sequences are balanced.
Conversely: say we have a balanced sequence of $1$, $-1$. Consider the first appearance of $-1$ from the left. Before it, there must be $1$. So our sequence
is of the form $A, 1, -1, B$. It is easy to see that $A,B$ is again balanced.
Therefore: legal is equivalent to balanced.
Consider now a sequence with $\sum_1^{2n} a_i = 0$ that is not legal. Then there must exist $k$ such that $\sum_1^k a_i < 0$. Consider $k$ smallest with such property. Then $k=1$, or $\sum_1^{k-1} a_i \ge 0$. The cannot be $> 0$, since then $a_k\le -2$. Therefore we have $\sum_1^{k-1}a_i = 0$, and $a_k=-1$ $\square$
Best Answer
This is the reflection principle in disguise.
You want to go from $(0,0)$ to $(n,n)$ on a grid in $n$ steps up and $n$ steps to the right without ever going below the diagonal, that is, without touching the line $L$ given by $y=x-1$. There are $\binom{2n}n$ such paths if we ignore the constraint. For paths that do touch $L$, reflect the part of the path up to the point where it first touches $L$ in $L$. This maps $(0,0)$ to $(1,-1)$. Each path from $(1,-1)$ to $(n,n)$ in $n+1$ steps up and $n-1$ steps to the right arises in exactly one way by this operation (since such a path necessarily at some point touches $L$). Thus there is a bijection between the paths from $(0,0)$ to $(n,n)$ that violate the constraint and the paths from $(1,-1)$ to $(n,n)$; and there are $\binom{2n}{n-1}$ of these paths.
To transfer this to the parenthesis paradigm, find the first closing parenthesis that makes the string unbalanced and toggle it and all parentheses before it (both opening and closing) to the other kind. This produces a string with $n+1$ opening and $n-1$ closing parentheses, and each such string is produced once. (I don’t know why the proof you have has it the other way around; you can instead toggle the parentheses after the first one that violates the constraint in order to get a bijection to strings with $n+1$ closing and $n-1$ opening parentheses.)