Does $a\mid bc$ imply $a\mid b$ or $a\mid c$?

In fact, your characterization is how primes are usually defined in algebra:

A prime is a number $1<p \in \Bbb Z$ such that for $a,b \in \Bbb Z$ with $p \mid ab$ we can conclude $p\mid a$ or $p \mid b$.

Not if $a$ is not prime. Consider $a=6$, $b=10$, $c=9$. Clearly $a$ divides $bc$ but not $b$ or $c$. This is because $\gcd(a,b) =2$ and $\gcd(a,c)=3$. So if we write $\frac{b\,c}{a}$ some factor of $a$ divides $b$ and the remaining factor divides $c$.

However if we can show that $a$ is coprime to one of $b$, $c$ then $a$ has to divide the other. Example $a=6$, $b=77$, and $a | b c$. We can say that $a$ has to divide $c$.

Proof: We know that if $d=\gcd(a,b)$, then we can find $u$ and $v$ so that $$ a u + b v = d $$ Now take the case where $\gcd(a,b)=1$. Then $$ u a + v b = 1 \Rightarrow u a c + v b c = c \tag 1 $$ If $a$ divides $b c$ then $$ b c = e a$$ Substituting in (1) we have $$ c = u a c + v b c = u a c + v e a = a \,(u c + v e) \Rightarrow a | c$$

Note that if $a$ is a prime number then either $a$ divides $b$ or $\gcd(a,b) = 1$. So we can conclude

If $a$ is prime and divides $bc$ then $a$ divides either $b$ or $c$ (or both)

HINT : Consider the case $(a,b,c)=(4, 6, 14)$.

Hint $\ bc\mid bc\ $ but neither $\,bc\mid b\,$ nor $\,bc\mid c,\,$ assuming $\,b,c > 1.$

Thus every composite $\,a = bc\,$ yields a counterexample. Conversely, it is true for prime $\,a\,$ by uniqueness of prime factorizations (or equivalent properties such as Euclid's Lemma), e.g. see here.

Remark $ $ If $\,a\,$ fails to have this property (i.e. Euclid's Lemma fails) then we can constructively prove $\,a\,$ is composite by explicitly constructing a proper factor of $\,a\,$ by a gcd calculation - see here.

Read the Euclid's Lemma and its generalization: http://en.wikipedia.org/wiki/Euclid's_lemma#Formulations

If $a$ is prime then what you wrote is true. And if gcd(a,b)=1 then a|c. Otherwise not. Consider a=6, b=2, c=3