Let $X, Y \in L^2$ on the probability space $(\Omega, \mathcal A, P)$ with the sub-$\sigma$-algebra $\mathcal B \subseteq \mathcal A$.
How can I show $\mathrm {Var}(X)=\mathrm E(X|\mathcal B)+\mathrm E[X-\mathrm E(X|\mathcal B)]^2\Rightarrow \mathrm {Var}(E(X|\mathcal B)))\le\mathrm {Var}(X)$
I am not sure what you meant in your expression for $Var[X]$. But to show $Var[E[X|\mathcal{B}]]\leq Var[X]$ you can use the law of total variance. That is $$Var[X]=Var[E[X|\mathcal{B}]]+E[Var[X|\mathcal{B}]].$$ The above implies $$Var[X]\geq Var[E[X|\mathcal{B}]] \iff E[Var[X|\mathcal{B}]]\geq 0.$$
Now to show the latter, let us write the definition of conditional variance: $Var[X|\mathcal{B}] = E[(X-E[X|\mathcal{B}])^2|\mathcal{B}]$ which is the expectation of a non-negative random variable and hence must be $P$-a.s. non-negative. Hence, $Var[X|\mathcal{B}]$ is a $P$-a.s. non-negative random variable which implies that its expectation must be non-negative.
If you were not allowed to use the law of total variance I provide you here with a proof of it: On one hand, $$Var[E[X|\mathcal{B}]]=E[E(X|\mathcal{B})^2]+E[E[X|\mathcal{B}]]^2 = E[(X|\mathcal{B})^2]+E[X]^2.$$ On the other hand, $$E[Var[X|\mathcal{B}]]=E\left[ E[(X-E[X|\mathcal{B}])^2|\mathcal{B}]\right]=\dots=E[X^2]-E[E[X|\mathcal{B}]^2].$$ Then both together gives you the desired identity.