Here is an argument using the well-ordering theorem, and no ordinals, which I find fairly intuitive (although some of this intuition probably does come from intuition about ordinal arithmetic). I spell a lot of facts about well-orders out, so if you know those facts already then there will be sentences and paragraphs that you can skip.
Let $\le$ be a well-ordering on $X$.
Firstly, we can assume that $X$ has no largest element. This is because any well-ordering can be (uniquely) decomposed as $X' + F$, where $F$ is finite and $X'$ has no largest element. This notation means that $X'$ and $F$ are well-ordered sets, and $X$ looks like "a copy of $X'$ followed by a copy of $F$". (Proof - if there is a largest element, let $x$ be the smallest element such that $\{y \in X : y \ge x\}$ is finite, and take $X' = \{y \in x : y < x\}$, $F = \{y \in X : y \ge x\}$).
So replace $X$ by the well-ordered set $F + X'$ if needs be (which is obviously the same size as $X' + F$). $X'$ must be non-empty, because $X$ was assumed to be infinite, so $F + X'$ has no largest element, because $X'$ has no largest element.
Now we need to show that $\{0, 1\} \times X$ bijects with $X$.
Note that $\{0, 1\} \times X$ is well-ordered by the colexicographic order induced by the usual order on $\{0, 1\}$ and the order $\le$ on $X$. If you think of $X$ as a discrete line of points, this order looks like what you get if you replace every point of $X$ by a copy of $\{0, 1\}$, or in other words, two points. Now I claim that this ordering is actually isomorphic to $(X, \le)$. Hopefully if you think about $X = \Bbb N$, for example, this is believable - in that case, having countably many copies of $\{0, 1\}$ next to each other is obviously the same order type as $\Bbb N$.
In fact, I think the nicest way to prove this conceptually is to prove that whenever $X$ is a well-ordering with no greatest element, then we can decompose $X$ as $\Bbb N \times X''$ for some well-ordered set $X''$ (think of this as saying that $X$ is a number of copies of $\Bbb N$).
Proof: take $X''$ to be the subset of elements of $X$ that have no direct predecessor in the ordering on $X$. Then the map $\Bbb N \times X'' \to X$ which sends $(n, x)$ to the $n$th successor of $x$ is an order isomorphism. (Take $0$ to be in $\Bbb N$ here).
It's well-defined, since $X$ has no largest elements.
It's strictly order-preserving, because if $x_1 < x_2$ have no direct predecessors, then the $n$th successor of $x_1$ is always less than $x_2$, by induction on $n$, and clearly if $n < m$ then the $n$th successor of $x_1$ is less than the $m$th successor of $x_1$. Hence it is also injective.
It's surjective, because any element of $X$ has only finitely many direct predecessors (otherwise the infinite sequence of direct predecessors would be a set with no smallest element).
You can think of this as some sort of transfinite exhaustion process - any well-order with no largest element starts with a copy of $\Bbb N$. Then after that, there is another copy, and another copy, ...
Given this, it's fairly clear that we can just use the bijection between $\{0, 1\} \times \Bbb N$ and $\Bbb N$ in each copy of $\Bbb N$. More formally, we have a chain of isomorphisms of ordered sets:
\begin{equation*}
\{0, 1\} \times X \cong \{0, 1\} \times (\Bbb N \times X'') \cong (\{0, 1\} \times \Bbb N) \times X'' \cong \Bbb N \times X'' \cong X.
\end{equation*}
So certainly $\{0, 1\} \times X$ and $X$ biject. (Here I used the fact that products of isomorphic total orders are isomorphic, and taking products of total orders is associative up to isomorphism, which are both straightforward.)
You could also have argued directly about cardinalities at the end without even worrying about the ordering anymore:
\begin{equation*}
|\{0, 1\} \times X| = |\{0, 1\} \times (\Bbb N \times X'')| = |(\{0, 1\} \times \Bbb N) \times X''| = |\Bbb N \times X''| = |X|.
\end{equation*}
This argument uses similar facts, but you only care about bijections, so they're very slightly more straightforward to prove.
As I mentioned in the comments, this argument is basically a translation of the following:
Choose some limit ordinal $\alpha$ in bijection with $X$. By Euclidean division, we can write $\alpha = \omega \beta$. Hence $\{0, 1\} \times X$ bijects with $2 \cdot \alpha = 2 \cdot \omega \beta = \omega \beta = \alpha$.
Alternatively, one can show by induction that whenever $\alpha$ is a limit ordinal and $\gamma < \alpha$, we have $2 \gamma < \alpha$, from which it follows that $2\alpha \le \alpha$.
Proof: If we know $2\gamma < \alpha$, then $2(\gamma + 1) = 2\gamma + 2 < \alpha$ because $\alpha$ is a limit. Also, if $\gamma < \alpha$ is a limit, then by induction on $\alpha$, we know that $2\delta < \gamma$ for all $\delta < \gamma$. Hence $2\gamma \le \gamma < \alpha$.
To prove the full result that $|X \times X| = |X|$ the argument is quite similar in spirit - you just define a well-ordering on the set $X \times X$ which makes it order-isomorphic to $X$. In that case, the order theory is more important, and so is the specific well-ordering that you choose on $X$ - you have to pick a well-ordering such that for every $x \in X$, the set $\{y \in X: y < x\}$ does not biject with $X$ ("an initial ordinal"). We may assume this because if there is an $x$ violating this, then we can consider the least such $x$ and just replace $X$ by the set $\{y \in X : y < x\}$, which bijects with $X$, but no proper initial segment of which bijects with $X$. This is a much, much stronger property than having no largest element.
Working with this well-ordering can also give an argument that shows $\{0, 1\} \times X$ bijects with $X$, by showing that each of its initial segments embeds into $X$. This does requires a fair bit more order theory, which I think is best understood by learning about ordinals. At that point you also might as well do the full proof for $X \times X$, after which you can easily apply Schröder-Bernstein by injecting $\{0, 1\} \times X$ into $X \times X$.
You can also use Zorn's lemma in various clever, but I think less enlightening ways. For example, the result follows from this already.
