Show that $X\subset X\cup\left\{ X\right\}$ Show that If $X$ is an any set, then $X\subset X\cup\left\{ X\right\}$


Proof. Let $t\in X$. We must show $t\in X\cup\left\{ X\right\}$, that is we need to show either $t\in X$ or $t\in\left\{ X\right\}$, so we know that $t\in X$, hence we are done.

Can you check my proof?
Apparently, an answer on ground level of set theory is needed here. Since the inclusion $X \subseteq X \cup \{X\}$ is already discussed in full detail, I want to make a remark about the case $X \neq X \cup \{X \}$.

The Axiom of Regularity reads
$$ \forall x \left( x \neq \emptyset \Rightarrow \exists y \in x: y \cap x = \emptyset \right).$$

Let us prove, that $X \notin X$ for all sets $X$.

Soo, let $X$ be any set. Due to the axiom of pairing, $\{X\}$ is a set aswell and clearly not empty (well, $X$ is an element). But in consequence of the axiom of regularity, we must have $X \cap \{X\} = \emptyset$ since $X$ is the only element in $\{X\}$. Hence, $X \notin X$.
Exercise.
Show for all sets A and B, that A $\subseteq$ A $\cup$ B.  

With that, your problem is just a special case.