Q:

Find a bijection between two sets

Show that there is a bijection mapping between $X^{\Bbb N}$ (sequences of elements of $X$) and a proper subset of itself.

A:

In this answer I will assume $0\in \mathbb{N}$.
Fix some elment $a\in X$ and take $\tilde X= \{a\}\times X^{\mathbb{N} -\{0\}}$, that is all the sequences (=elements of $X^\mathbb N$) that start with $a$.
You can find a bijection between $\tilde X$ and $X^\mathbb N$, say $(x_0,x_1,\ldots) \mapsto (a,x_0,x_1,\ldots)$.
You can also show $|\tilde X|=|X^\mathbb N|$ and deduce that such bijection exists, without explicitly finding it.
Please note that this holds iff $X\not=\{a\},\emptyset$, that is $X$ has more than one element. If it has exactly one element, then $|X^\mathbb N|=|(a,a,a,\ldots)|=1$ and clearly the proposition is not true (in that case, $\tilde X$ is not a proper subset of $X^\mathbb N$) . Also, the same hapens if $X=\emptyset$ (in that case, you can't define $\tilde X$ the way I did as there is no $a\in X$). So we need $|X|\geq 2$ for this argument.

