Question: What I want to do is to mathematically prove the following two inequalities: $$Var(Y) \geq Var(Y | X)$$ $$Var(Y | X_1) \geq Var(Y | X_1, X_2)$$
Some hint to the proof will be appreciated. Thanks in advance!
Details Let $X_1, X_2$, and $Y$ be continuous random variables. We can also assume that they are not independent from each other.
This is not true. (More precisely, this equality does not really make sense, at least not the way you most likely intended.)
By definition, $$\begin{align} \operatorname{Var}(Y\mid X) &= \mathbb{E}[(Y - \mathbb{E}[Y\mid X])^{2}\mid X] = \mathbb{E}[Y^2 - 2Y\mathbb{E}[Y\mid X]+\mathbb{E}[Y\mid X]^{2}\mid X]\\ &= \mathbb{E}[Y^2\mid X] - 2\mathbb{E}[Y\mathbb{E}[Y\mid X]\mid X]+\mathbb{E}[\mathbb{E}[Y\mid X]^{2}\mid X]\\ &= \mathbb{E}[Y^2\mid X] - 2\mathbb{E}[Y\mid X]^2+\mathbb{E}[Y\mid X]^{2}\\ &= \mathbb{E}[Y^2\mid X] - \mathbb{E}[Y\mid X]^2 \end{align}$$ expanding and using the fact that $\mathbb{E}[Y\mid X]$ is $\sigma(X)$-measurable.
Note that this is not a constant, but a random variable, so intuitively the inequality you seek is a "type error."
To see a concrete example, consider the simplest case, when $Y$ has an arbitrary distribution (to be picked later) and $X=Y^2$:, say for simplicity $Y=RN$ where is a Rademacher random variable (uniform on $\{-1,1\}$) independent of $N$ which is Gaussian$(0,1)$.
Then $\mathbb{E}[Y^2\mid X] = \mathbb{E}[X\mid X] = X$, $\mathbb{E}[Y\mid X]^2 = \mathbb{E}[Y\mid Y^2]^2 = \mathbb{E}[RN\mid N^2]^2= \mathbb{E}[R]\mathbb{E}[N\mid N^2]^2=0$ so that $\operatorname{Var}(Y\mid X) = X$.
But $\operatorname{Var}(Y)$ is just a nice number, and since $X$ is continuous and supported on $\mathbb{R}$ we have $\mathbb{P}\{X > \operatorname{Var}(Y) \} > 0$.
That is,
By the law of total variance, however, you do have: $$ \operatorname{Var} Y = \mathbb{E}[\operatorname{Var}( Y \mid X)] + \underbrace{\operatorname{Var}( \mathbb{E}[Y\mid X])}_{\geq 0} \geq \mathbb{E}[\operatorname{Var}( Y \mid X)]. $$ Note that here both LHS and RHS are "numbers."