I suggest the following counter-example.

Theorem. In a (finite) color pencils box, all pencils must have the same color.

Proof. We proceed by induction on the number of pencils. If the box has one pencil, then obviously all its pencils have the same color. Now, assume that in all boxes of $n$ pencils, the pencils are all of the same color and let $X$ be a box with $n+1$ pencils. Order the pencils arbitrarily and consider the $n$ first pencils: by the induction hypothesis, they all have the same color, call it $c$. Consider now the $n$ last pencils: by the induction hypothesis, they all have the same color, call it $c'$. Since the $n-1$ central pencils have color $c$ and $c'$, we have $c=c'$ so that all $n+1$ pencils have the same color.

The goal here is to stress that if one fails to properly identify and test the base case, then things can go dead wrong.

I suggest the following counter-example. (Added in edit: I agree with kcrisman comment that this may be misleading to beginners, and should probably be kept for people that master induction relatively well and start being lazy verifying the base case).

Theorem. In a (finite) color pencils box, all pencils must have the same color.

Proof. We proceed by induction on the number of pencils. If the box has one pencil, then obviously all its pencils have the same color. Now, assume that in all boxes of $n$ pencils, the pencils are all of the same color and let $X$ be a box with $n+1$ pencils. Order the pencils arbitrarily and consider the $n$ first pencils: by the induction hypothesis, they all have the same color, call it $c$. Consider now the $n$ last pencils: by the induction hypothesis, they all have the same color, call it $c'$. Since the $n-1$ central pencils have color $c$ and $c'$, we have $c=c'$ so that all $n+1$ pencils have the same color.

The goal here is to stress that if one fails to properly identify and test the base case, then things can go dead wrong.