This identity can be visualized in a Pascal triangle. The two terms on the left are two neighboring terms on the $n^{th}$ row which form two shoulders. The term on the right side is the term beneath the two shoulders.

It can also be proved using the basic definition. $$\begin{array}{rl} \binom{n}{k} + \binom{n}{k+1} =& \frac{n!}{k!(n-k)!}+\frac{n!}{(k+1)!(n-k+1)!} \\ =&\frac{n!(k+1)}{(k+1)!(n-k)!} + \frac{n!(n-k)}{(k+1)!(n-k)!} \\ =&\frac{(n+1)!}{(k+1)!(n-k+1)!}\\ =&\binom{n+1}{k+1}\end{array}$$