Unprovable means that we cannot prove it. Given a list of axioms, $\varphi$ is provable from these axioms if there is a finite sequence of statements all of which are axioms, or deduced from axioms and previously written statements in the sequence, such that the final statement in the sequence is $\varphi$.
So when is $\varphi$ unprovable? When such finite sequence doesn't exist. How do we prove that something doesn't exist? In particular something like that?
Luckily we know that if $\sf ZF$ proves $\varphi$ then in every model of $\sf ZF$ the statement $\varphi$ must be true, on the other hand if we can only show there is some models of $\sf ZF$ where $\varphi$ is true then $\varphi$ is consistent with $\sf ZF$, being provable implies being consistent with, but usually not vice versa (in particular in the case of $\sf ZF$). So if we can find two models of $\sf ZF$, one where $\varphi$ is true and another where it's not, then we show that $\varphi$ is not provable from $\sf ZF$.
This is how the continuum hypothesis was proved to be unprovable. If $\sf ZF$ is consistent (otherwise it proves everything anyway), then it has a model where $\sf CH$ is true, this was shown by Goedel in the 1940's. Some 20 years later Cohen showed how to construct from Goedel's model a model of $\sf ZF$ where the continuum hypothesis is false. This construction is technical and requires more than basic understanding of set theory, so I won't get into it right now.
But all combined we have that if $\sf ZF$ is consistent to begin with, then it is consistent with the continuum hypothesis and with its negation. Meaning that the continuum hypothesis is unprovable from $\sf ZF$.