I made a post earlier just focusing on the injective case but now I've extended this to bijectivity:
Injectivity:
By injectivity (this isn't neccesary to write but I put it here for reference):
$g(x)=g(x')\implies x=x'$
$f(x)=f(x')\implies x=x'$
$(g\circ f)(x)=(g\circ f)(x')\implies f(x)=f(x')\implies x=x'$
Therefore $g\circ f$ is injective.
Surjectivity:
Here is where I get confused, I usually get lost when it comes to surjectivity.
Suppose $f: X \rightarrow Y, g:Y\rightarrow Z$
For $f(x)$: $\forall y\in Y, \exists x\in X:f(x)=y$
For $g(y)$: $\forall z\in Z, \exists y\in Y: g(y)=z$
For $g(f(x))$: $\forall z\in Z, \exists x:\exists y:g(y)=z$. In other words, for any $z$, there is a $y$ that satisfies $g(y)=z$ since for any $y$ there is an $x$ that satisfies $f(x)=y$, thusly, surjectivity has occured.
Therefore $g \circ f$ is surjective, and thus it is bijective.
Best Answer
Deep Breath: And do it in the proper order.
For every $z\in Z$ we need to show there is an $x\in X$ so that $g(f(x)) = z$.
So we start in $Z$ and pull our way eventually to $X$ but go through $Y$ as the intermediate step.
So to go from $z \in Z$ what is it that gets up onto $Z$?
All this is me asking leading question to get you to note that we use $g:Y\to Z$ and so we start with $g$ being surjective. We do not start with $f$.
Now it's every bit as easy as they injective part!
$g:Y \to Z$ is surjective so for every $z\in \mathbb Z$ there is (at least) one $y \in \mathbb Y$ so that $g(y) =z$.
And $f: X\to Y$ is surjective so for the $y$ we used above (the specific $y$ so that $g(y) = z$, and not a generic $y$ in general) there is (at least) one $x\in X$ so that $f(x) = y$.
So for that $x$ we have $g(f(x)) = g(y) = z$ and for any $z$ we found at least one $x \in X$ so that $g\circ f(x) = z$.
So $g \circ f$ is surjective.