There is an inherent problem with your question. Actually two.
The first is that visualizing something requires some sort of inherent structure. We can visualize the real numbers as a line, or the natural numbers as dots, and $\Bbb R^3$ as the room you are sitting in.
When you talk about cardinality, you—by definition—omit the structure. Note that $\Bbb N$ and $\Bbb Z$ are very different, but they are both countable. Even $\Bbb N$ with the usual order, and with the divisibility relation ($m\mid n\iff\exists k\in\Bbb N:mk=n$) are very different visually, despite being two structures of the same set.
So can we come up with an easy to visualize structure? Sure. We have linear orders of any cardinality, as follows from the axiom of choice, so we can take one which "locally" looks like $\Bbb R$, or even like $\Bbb N$ (again, locally is a key word here). Does that mean something to you? How is that different from $\Bbb R^2$ viewed with the lexicographic order? It really doesn't.
The second problem is relying on "visualizing". Yes, you can visualize the function $f(x)=x$ or $f(x)=e^x$. If you try very hard, you can even differentiate in your mind's eye between $\ln x$ and $\sqrt x$ (but I can't, their graphs are just too similar). Mathematics is built on definitions. You could argue, and correctly so, that $\Bbb Q$ and $\Bbb{R\setminus Q}$ are extremely different objects. While both are linearly ordered metric spaces, only one of them is completely metrizable, only one of them is a $G_\delta$ subset of $\Bbb R$, only one of them is uncountable, and only one of them is a field. Even though, they are both totally disconnected and without isolated points.
So if you think about either $\Bbb Q$ or $\Bbb{R\setminus Q}$ visually, you are likely to think about a line with "missing dots almost everywhere", but also dots are still everywhere". Sort of to signify that both the set and its complement in the real line are dense and co-dense sets. And therein lies the rub. We just talked about how different these mathematical objects are, from several important and distinctive perspectives. Yet, visually speaking, they "look and feel the same".
This is why visualizing is not a good direction. It is not a good motivating tool. Abstract mathematics, and set theory in particular, is prone to confuse you if you try to visualize things as a crutch and a scaffolding device. You should rely on the definitions you have, and work carefully. With time you will gain the proper intuition for handling definitions related to infinite sets and their cardinality. It's not the easy way, but you won't be cheating yourself with the idea that you can somehow visualize well enough to make the distinction between two objects which are very different from one another.
For the first one, note that there is a surjection from $X$ onto $X/{\equiv}$. This means that the fibres of this functions define an injection from $X/{\equiv}$ into $2^X$, but you can show that this injective function actually maps $2^{X/{\equiv}}$ into $2^X$ injectively.
For the second one, note that quite literally, $X/{\equiv}$ is a subset of $2^X$, when considering $2^X$ as the power set of $X$. So we trivially get $\leq$. But since there is a surjection from $X$ onto $X/{\equiv}$, this $\leq$ cannot possibly be $=$, since that would contradict Cantor's theorem, which holds without choice (there is no surjection from $X$ onto $2^X$). Therefore $X/{\equiv}<2^X$.
Best Answer
This can't be proved without the axiom of choice. Or, more precisely, it cannot be proved in ZF alone.
It is known to be consistent with ZF that an amorphous set exists -- meaning an infinite set that is not the disjoint union of two infinite sets. Such a set would be an infinite set whose cardinality is incomparable with that of $\mathbb N$ and therefore be a counterexample to the claim.
(You don't need the full axiom of choice, though. The naive pick-a-sequence argument will go through with the axiom of dependent choice, and even the axiom of countable choice is enough, with a slightly more involved argument).