How do I rigorously show $f(\cap \mathcal{C}) \subset \cap f(\mathcal{C})$ I've already proved that $f(\cap \mathcal{C}) \subset \cap f(\mathcal{C})$. I'm going to mark numbers because they will be useful for my questions.
$\Rightarrow$

Let $\mathcal{C}$ be a collection of sets $C_n \subset \mathcal{C}$.
Let $y \in f(\cap \mathcal{C})$.
$(1)$ Then $x \in \cap \mathcal{C}$.
Then $\forall C \subset \mathcal{C}$ we have $x \in C$.
So $\forall C \subset \mathcal{C}$ we have $y \in f(\mathcal{C})$.
So $y \in \cap f(\mathcal{C})$.

$(1)$ First of all, if there exists a $f(x)$, does that imply there exists an $f^{-1}(f(x))$? Or, if there exists an $f^{-1}(f(x))$, does that imply there exists an $f(x)$? I think only the latter is true because the existence of an image does not imply the existence of a pre-image, but the existence of a pre-image does imply the existence of an image.
$(2)$ THE MOST IMPORTANT QUESTION: The converse would follow directly but that would show $=$, not $\subset$. So would I be assuming something incorrectly?
 A: For your first question, if you have an element in the image under a function, there must be something that mapped to it (otherwise it wouldn't be in the image). Perhaps you're confusing this with taking an arbitrary element in the range; this indeed does not necessarily have a preimage (take $f:\Bbb R\to\Bbb R$ defined by $f(x)=\sin(x)$ for example, with $y=2$). Also beware that the preimage of a single point may not consist if a single point (in the same example, what is $f^{-1}(1)$?).
To see the converse fail, take the following example: $\Bbb R=\Bbb Q\cup \Bbb Q^c$ and define $f$ to be your favorite constant function; I'll say $f(x)=0$ for all $x$. Then the left-hand side of your containment is the empty set, while the right-hand side is $\{0\}$.
A: Here is a complete proof which may help explain why only one of the directions is true in general.
But first, let's define slightly different notations to prevent confusion between expressions like $\;f(x)\;$, $\;f(C)\;$, and $\;f(\mathcal{C})\;$: we write the last two as $\;f[C]\;$ and $\;f[\![\mathcal{C}]\!]\;$, respectively, so that we have the following definitions:
\begin{align}
(0) \;\;\; & y \in f[V] \;\equiv\; \langle \exists x : f(x) = y : x \in V \rangle \\
(1) \;\;\; & W \in f[\![\mathcal{B}]\!] \;\equiv\; \langle \exists V : f[V] = W : V \in \mathcal{B} \rangle \\
\end{align}
With these notations, the statement to prove is
$$
f[\cap \mathcal{C}] \;\subseteq\; \cap f[\![\mathcal{C}]\!]
$$
Our strategy is to start at (what I think is) the most complex side, the right hand side, calculate all its elements $\;y\;$ by expanding the definitions, and then seeing where the laws of logic lead us.

For all $\;y\;$,
\begin{align}
& y \in \cap f[\![\mathcal{C}]\!] \\
\equiv & \qquad \text{"definition of $\;\cap\;$"} \\
& \langle \forall W : W \in f[\![\mathcal{C}]\!] : y \in W \rangle \\
\equiv & \qquad \text{"definition $(1)$"} \\
& \langle \forall W : \langle \exists V : f[V] = W : V \in \mathcal{C} \rangle : y \in W \rangle \\
\equiv & \qquad \text{"logic: rewrite $\;\exists V\;$ in range of $\;\forall W\;$ to a $\;\forall\;$"} \\
& \langle \forall W,V : f[V] = W \land V \in \mathcal{C} : y \in W \rangle \\
\equiv & \qquad \text{"logic: one-point rule for $\;W\;$"} \\
& \langle \forall V : V \in \mathcal{C} : y \in f[V] \rangle \\
\equiv & \qquad \text{"definition $(0)$"} \\
& \langle \forall V : V \in \mathcal{C} : \langle \exists x : f(x) = y : x \in V \rangle \rangle \\
(*) \;\;\; \Leftarrow & \qquad \text{"logic: $\;\exists\forall \Rightarrow \forall\exists\;$} \\
& \qquad \phantom{\text{"}}\text{-- without any knowledge about $\;f\;$ or $\;\mathcal{C}\;$,} \\
& \qquad \phantom{\text{"-- }}\text{this is the only thing we can do;} \\
& \qquad \phantom{\text{"-- }}\text{this is where $\;\cap\;$ and applying $\;f\;$ are exchanged"} \\
& \langle \exists x : f(x) = y : \langle \forall V : V \in \mathcal{C} : x \in V \rangle \rangle \\
\equiv & \qquad \text{"definition of $\;\cap\;$"} \\
& \langle \exists x : f(x) = y : x \in \cap \mathcal{C} \rangle \\
\equiv & \qquad \text{"definition $(0)$"} \\
& y \in f[\cap \mathcal{C}] \\
\end{align}
By the definition of $\;\subseteq\;$, this completes the proof.

Note that the key step $(*)$ cannot be reversed in general because of the essentially one-directional law $\;\exists\forall \Rightarrow \forall\exists\;$.
Finally, looking at the proof as a whole, I am pleasantly surprised by the simple structure: until the key step $(*)$ we are only expanding definitions and simplifying, and after that point we immediately go back to the set and function level using the definitions.
