Consider a vector x in $R^n$ and a subspace V of $R^n$. Then the orthogonal projection $proj_V x$ is the vector in V closest to x, in that $||x-proj_V x||<||x-v||$ for all v in V different from $proj_V x$.
How to understand this theorem intuitively? and this is the image I found related to this theorem. How to prove it? Pythagoran theorem? 
Pythagoras's theorem is indeed a good way of thinking about this. If $\vec{w} = \operatorname{proj}_V \vec{x}$, then given any $\vec{v} \in V$, the vector $\vec{w} - \vec{v}$ lies in $V$ (after all $V$ is a subspace). The definition of the orthogonal projection is such that $\vec{x} - \vec{w}$ is orthogonal to every vector in $V$, so by Pythagoras's theorem, $$\|\vec{x} - \vec{v}\|^2 = \|\vec{x} - \vec{w}\|^2 + \|\vec{w} - \vec{v}\|^2 \ge \|\vec{x} - \vec{w}\|^2,$$ with equality if and only if $\|\vec{w} - \vec{v}\|^2 = 0$, i.e. $\vec{v} = \vec{w} = \operatorname{proj}_V \vec{x}$.