Regular induction ("holds for $1$" and "if it holds for $k$ then it holds for $k+1$") only gives you that the result holds for every natural number $n$; it does not let you go beyond the finite numbers. For example, you can prove by induction that there are natural numbers that require $n$ digits to write down in base $10$ for every $n$, but this does not mean that there are natural numbers that require an infinite number of digits to write down in base $10$. "For all $n$" is not the same as "for all sizes, finite or infinite".
(There is a kind of induction that would allow you to prove something for all sizes, not just finite. This is called transfinite induction. You prove the result holds for $1$, and that whenever it holds for all $m\lt k$, then it also holds for $k$ (or, you prove it holds for $1$, that if it holds for an ordinal/cardinal $\alpha$ then it holds for $\alpha+1$, and that if it holds for all ordinals/cardinals strictly smaller than $\gamma$, then it holds for $\gamma$). However you would be unable to do such a proof with lattices, because it is false).
So, if you have a lattice, then any nonempty finite subset has a least upper bound and a greatest lower bound, by induction. Even if you have a $0$ and a $1$ (a minimum and a maximum element) so that every set has an upper and a lower bound, you still don't get that every set has a least upper bound. For example, take $P = \mathbb{Q}\cup\{-\infty,\infty\}$, with the usual order among rationals, $-\infty\leq q\leq \infty$ for all $q\in\mathbb{Q}$. This is a lattice, with operations $a\wedge b = \min\{a,b\}$ and $a\vee b = \max\{a,b\}$ (since it is a totally ordered set). Every finite subset has a least upper bound (the maximum) and a greatest lower bound (the minimum). But it is precisely the absence of suprema and infima for general sets that stops it from being a complete lattice: the set $\{q\in\mathbb{Q}\mid q^2\lt 2\}$ has no least upper bound and no greatest lower bound.