Finding an example of a non-rational p-adic number We know that every rational number can be written as a $p$-adic integer with expansion $\sum\limits_{n=-m}^\infty a_n p^n$, where $a_n\in\\{0,\dots,p-1\\}$ and $m\in\mathbb{N}$; therefore there exists an injection $\mathbb{Q} \hookrightarrow \mathbb{Q}_p$. But how do I show that $\mathbb{Q}_p$ is bigger? How do I find an example of a $p$-adic number which is not rational? I heard $p$-adic numbers and even $p$-adic integers being uncountable, is the proof easy?

First of all, it is every integer (not every rational) that can be written as you have presented it as a sequence $a_n\in\\{0,\ldots, p-1\\}$. The number $1/p$ cannot be so represented.

To answer your cardinality question, there are at least two arguments that $\mathbb Z_p$ is uncountable, and thus that $\mathbb Q_p$ is uncountable.

From your presentation of the $p$-adics as the series $\sum a_n p^n$, we see that the $\mathbb Z_p$ has the cardinality of the set of maps $\mathbb N^+\rightarrow \\{0,\ldots, p-1\\}$, (i.e., $n\mapsto a_n$) and hence uncountable.

Another argument is to start with the fact that $\mathbb Z_p$ is infinite and (Hausdorff) compact, e.g., because $\mathbb Z_p= \lim \mathbb Z/p^{n}$, so a closed set of a compact set, hence compact (and infinite). By Baire's theorem, in a compact Hausdorff space, the countable union of nowhere dense closed sets cannot contain a non-empty open set. In particular, an infinite countable set cannot be (Hausdorff and) compact.