Given a rational $\succsim$ over a finite set $X$, show that there exists $x \in X$ such that $x \succsim y, \forall y \in X$ I have been able to show this constructively, but would like to prove it by induction. However, I am stuck with the induction step:

Consider $\succsim$ defined over $X=\{x_1,...,x_n\}$ and where without loss of generality $x_1$ is the element such that $x_1 \succsim x, \forall x \in X$. Now, consider adding element $x_{n+1}$ to the set. My reasoning is that since $\succsim$ is defined over the original $X$, there is nothing telling us that $x_{n+1}$ should be in a complete relation with the elements in $X$ (i.e. $\succsim$ is not defined over a bigger set).

Is this reasoning correct? In that case, I don't see how you could prove this by induction.
You are almost there, but I think you are starting in the wrong place. I think the key is as follows: we want to prove the claim for set $X=\{x_1,\ldots,x_n\}$. 

Start with some $X'\subseteq{X}$ for which the claim is true. Such an $X'$ exists because the claim is true for the set $\{x_1\}$ (by completeness and reflexivity). 

If $X'=X$ then we're done. If $X'\neq X$ then we proceed with induction:

Denote by $\bar{x}$ an $x\in X'$ such that $\bar{x}\succsim y, \forall\ y\in X'$.

Now construct a new set $X''=X'\cup\{x_i\}$, where $x_i\in X\backslash X'$. Because $\succsim$ is rational (preferences are complete) over $X$, we either have $x_i\succsim \bar{x}$ or $\bar{x}\succsim x_i$ or both. Thus, if the claim holds for $X'$ it also holds for $X''$.

We can repeat this inductive step until $X''=X$, which must happen after finitely many iterations because $X$ is finite.