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Of course, the comment by Jim Humphreys and the answer by Marty Isaacs are are entirely accurate. Using the original character-theoretic definition of defect of a block, introduced by Brauer, the existence of a height zero character is almost tautological. However, there are ring-theoretic definitions of defect of a block which may make the existence of height zero characters les obvious. I illustrate with an example: If the finite group $G$ has a Sylow $p$-subgroup $P$ of order $p^{a},$ then there is a fairly short ring-theoretic proof by J.A. Green that the dimension of a block $B$ of defect $d$ of $FG$ ($F$ algebraically closed of characteristic $p$) is divisible by $p^{2a-d}$ but by no higher power of $p.$ ( This result had been proved previously by Brauer using characters). But the dimension of $B$ is also equal to $\sum_{S} {\rm dim}(P(S)) {\rm dim}(S),$
where $S$ runs through the simple $FG$-modules in $B$ and $P(S)$ denotes the projective cover of $S$. Each $P(S)$ has dimension divisible by $p^{a},$ as $P(S)$ is projective, and each ${\rm dim S}$ is divisible by $p^{a-d},$ (which can be proved in a ring-theoretic fashion via Green's theory of vertices). Thus there is some simple $B$-module $S$ whose dimension is exactly divisible by $p^{a-d}.$ Since Brauer characters of simple $B$-modules are integral combinations of (restrictions to $p$-regular elements of ) ordinary irreducible characters in $B),$ it follows that there is an ordinary irreducible character in $B$ whose degree is divisible by $p^{a-d},$ but by no higher power of $p,$ that is, an irreducible character of height zero. My purpose is not to point out the shortest proof, merely to point out that it is possible to start from different viewpoints, and still see that there are characters of height zero in every block.