Can a function that selects between two random variables increase the variance more than twofold? Let $X_1,X_2$ be two real-valued zero-mean random variables, and assume w.l.o.g. that $\text{Var}[X_1]\ge\text{Var}[X_2]$.
Let $f:\mathbb R^2\to\{1,2\}$ be a ``selection'' function, and define $Y=X_{f(X_1,X_2)}$ to be the selected (real-valued) random variable.
Is it possible to upper bound $\text{Var}[Y]$ as a function of $\text{Var}[X_1]$? For example, 

Is it correct that $\text{Var}[Y]\le 2\text{Var}[X_1]$?

 A: Let $A_1, A_2$ be the disjoint events $\{f=1\}, \{f=2\}$.  Then we can write
$$Y = 1_{A_1} X_1 + 1_{A_2} X_2.$$
Note that
$$Y^2 = 1_{A_1} X_1^2 + 1_{A_2} X_2^2$$
Now
$$\begin{align*} \operatorname{Var}(Y) &\le E[Y^2] \\
&= E[1_{A_1} X_1^2] + E[1_{A_2} X_2^2] \\
&\le E[X_1^2] + E[X_2^2] \\
&= \operatorname{Var}(X_1) + \operatorname{Var}(X_2)  && \text{(since $E[X_1]=E[X_2]=0$)}\\
&\le 2 \operatorname{Var}(X_1).
\end{align*}$$
A: This is true (your conjecture about constant $k$ for $k$ variables is also true), and is a consequence of law of total variance. More specifically, this is the law of total variance
$ Var(Y) = \mathbb{E}[Var(Y | F)] +Var(\mathbb{E}[Y \mid F])$
where I am denoting $f(X_1, X_2)$ as $F$.
Further simplification gives
$ Var(Y) \leq \sum_i Var(X_i \mid f = i) P(f = i)  + \mathbb{E}[X_i \mid f = i]^2 P(f = i) $
which means
$Var(Y) \leq \sum_i \mathbb{E}[ X_i^2 \mid f = i] P(f = i) $
Now, law of total probability here, coupled with the fact that $X_i^2 \geq 0$ and $X_i$ have 0 mean, gives you what you want. 
EDIT to answer the comment:
$\sum_i \mathbb{E}[ X_i^2 \mid f = i] P(f = i) \leq \sum_i \mathbb{E}[ X_i^2 \mid f = i] P(f = i) + $
$\sum_{j \neq i} \mathbb{E}[ X_i^2 \mid f = j] P(f = j) = \sum_i \mathbb{E} X_i^2 = \sum_i Var(X_i) $
