Consider $X$ a countably infinite set, is $\{X \cup x\}$ with $x \notin X$ countably infinite? $x$ represents an element, not a set. 
Assume $S:=\{X \cup x\}$ is countably finite. This means that there is a bijection between $S$ and a subset of $\mathbb{N}$. This is clearly not the case since $X$ is infinite. I don't really know how to write down that last part rigorously. How can I approach this? 
Thanks in advance.
 A: As the comments point out, your notation is a little confusing. The set $S = \lbrace X\cup x\rbrace$ contains one element: $X\cup x$. In particular, $S$ is countable. On the other hand, the set $X\cup x$ may not be countable, even if $X$ is countable. This is because $x$ could be uncountable and disjoint from $X$. The statement you are probably looking for is the following: 
Proposition. If $X$ is countable and $x\not\in X$, then $X\cup \lbrace x\rbrace$ is countable. 
In fact, something much stronger is true. If $X_1,X_2,\dots $ is a sequence of countable sets (indexed over the natural numbers), then $\bigcup_i^\infty X_i$ is countable. To prove this, use that $\mathbb N\times\mathbb N$ is countable and construct a surjection $\mathbb N\times\mathbb N\rightarrow \bigcup_i^\infty X_i$. The proposition above is then the special case where $X_1 = X$, $X_2 = \lbrace x\rbrace$ and $X_i = \varnothing$ for $i > 2$. 
Addendum. Now how would you prove that $\mathbb N\times\mathbb N$ is countable? Here it may help to visualize $\mathbb N\times \mathbb N$ as the nodes of an infinite grid in the first quadrant of $\mathbb R^2$. A natural way to "count" these nodes is to "spiral outwards". 
