Prove the identity $\binom{2n+1}{0} + \binom{2n+1}{1} + \cdots + \binom{2n+1}{n} = 4^n$

Why not just $$ \begin{align} 2^{2n+1} &=\binom{2n+1}{0}+\cdots+\binom{2n+1}{n}+\binom{2n+1}{n+1}+\cdots+\binom{2n+1}{2n+1} \\ &=\binom{2n+1}{0}+\cdots+\binom{2n+1}{n}+\binom{2n+1}{n}+\cdots+\binom{2n+1}{0} \\ &=2\left[\binom{2n+1}{0}+\cdots+\binom{2n+1}{n}\right] \end{align} $$ Then divide each extremity by 2.

Applying what you mention on (in this context nicer) odd $2n+1$ instead of even $2n$ we find:$$2\times4^n=2^{2n+1}=\sum_{k=0}^{2n+1}\binom{2n+1}{k}=2\sum_{k=0}^{n}\binom{2n+1}{k}$$

Now divide by $2$.

Let $A$ be the powerset (i.e., the set of subsets of) of $\{1,\ldots,2n\}$. Let $B$ the set of all subsets of $\{1,\ldots,2n+1\}$ at most $n$ elements. Then "clearly" $|A|=2^{2n}=4^n$ and $|B|=\sum_{k=0}^n{2n+1\choose k}$. Define the following map $f\colon B\to A$: $$f(S)=\begin{cases}S&\text{if }2n+1\notin S\\\{1,\ldots,2n+1\}\setminus S&\text{if }2n+1\in S\end{cases} $$ and define $g\colon A\to B$ by $$g(S)=\begin{cases}S&\text{if }|S|\le n\\\{1,\ldots,2n+1\}\setminus S&\text{if }|S|>n\end{cases} $$ Finally, verify that $f$ and $g$ are inverse of each other, hence $|A|=|B|$.