An idea to start this inequality $$Var\left ( E\left [ Y \mid  X \right ] \right )\leqslant Var \left ( Y \right )$$
 A: We assume that $X$ and $Y$ are random variables on the same probability space, and the variance of $Y$ is finite.
We have that $\mbox{E}[\mbox{Var}[Y \mid X])]\geq 0$ because variance is nonnegative. Therefore by Law of total variance,
$$\mbox{Var}(Y) = \mbox{E}[\mbox{Var}[Y \mid X])] + \mbox{Var}[\mbox{E}[Y \mid X]]\geq \mbox{Var}[\mbox{E}[Y \mid X]].$$
A: In order to understand the reason behind, an explanation may be given in terms of orthogonal projections: Let  our probability space be $L^1(\Omega,{\cal A}, P)$ and let ${\cal B} \subset {\cal A}$ be the $\sigma$-algebra generated by $X$. Then  $E(Y|X)$ is the ${\cal B}$ measurable random variable obtained by orthogonal projection of $Y\in L^2(\Omega,{\cal A},P)$ onto $L^2(\Omega,{\cal B},P)$. Therefore, we may write
$$ Y = E(Y|X) + (Y-E(Y|X)), $$
where the two terms on the right are perpendicular (for the scalar product in $L^2(\Omega,{\cal A},P)$). The last term has zero average so also
  $$Y-E(Y) = (E(Y|X)-E(Y)) + (Y-E(Y|X))$$
is a sum of two orthogonal vectors. So by Pythagoras:
  $$ {\rm Var}(Y) = E ((E(Y|X)-E(Y))^2) + E((Y-E(Y|X))^2)={\rm Var} (E(Y|X))+ E ({\rm Var}(Y|X))$$
