Commuting matrices do not necessarily share all eigenvector, but generally do share a common eigenvector.
Let $A,B\in\mathbb{C}^{n\times n}$ such that $AB=BA$. There is always a nonzero subspace of $\mathbb{C}^n$ which is both $A$-invariant and $B$-invariant (namely $\mathbb{C}^n$ itself). Among all these subspaces, there exists hence an invariant subspace $\mathcal{S}$ of the minimal (nonzero) dimension.
We show that $\mathcal{S}$ is spanned by some common eigenvectors of $A$ and $B$.
Assume that, say, for $A$, there is a nonzero $y\in \mathcal{S}$ such that $y$ is not an eigenvector of $A$. Since $\mathcal{S}$ is $A$-invariant, it contains some eigenvector $x$ of $A$; say, $Ax=\lambda x$ for some $\lambda\in\mathbb{C}$. Let $\mathcal{S}_{A,\lambda}:=\{z\in \mathcal{S}:Az=\lambda z\}$. By the assumption, $\mathcal{S}_{A,\lambda}$ is a proper (but nonzero) subspace of $\mathcal{S}$ (since $y\not\in\mathcal{S}_{A,\lambda}$).
We know that for any $z\in \mathcal{S}_{A,\lambda}$, $Bz\in \mathcal{S}$ since $\mathcal{S}_{A,\lambda}\subset\mathcal{S}$ and $\mathcal{S}$ is $B$-invariant. However, $A$ and $B$ commute so
$$
ABz=BAz=\lambda Bz \quad \Rightarrow\quad Bz\in \mathcal{S}_{A,\lambda}.
$$
This means that $\mathcal{S}_{A,\lambda}$ is $B$-invariant. Since $\mathcal{S}_{A,\lambda}$ is both $A$- and $B$-invariant and is a proper (nonzero) subspace of $\mathcal{S}$, we have a contradiction. Hence every nonzero vector in $\mathcal{S}$ is an eigenvector of both $A$ and $B$.
EDIT: A nonzero $A$-invariant subspace $\mathcal{S}$ of $\mathbb{C}^n$ contains an eigenvector of $A$.
Let $S=[s_1,\ldots,s_k]\in\mathbb{C}^{n\times k}$ be such that $s_1,\ldots,s_k$ form a basis of $\mathcal{S}$. Since $A\mathcal{S}\subset\mathcal{S}$, we have $AS=SG$ for some $G\in\mathbb{C}^{k\times k}$. Since $k\geq 1$, $G$ has at least one eigenpair $(\lambda,x)$. From $Gx=\lambda x$, we get $A(Sx)=SGx=\lambda(Sx)$ ($Sx\neq 0$ because $x\neq 0$ and $S$ has full column rank). The vector $Sx\in\mathcal{S}$ is an eigenvector of $A$ and, consequently, $\mathcal{S}$ contains at least one eigenvector of $A$.
EDIT: There is a nonzero $A$- and $B$-invariant subspace of $\mathbb{C}^n$ of the least dimension.
Let $\mathcal{I}$ be the set of all nonzero $A$- and $B$-invariant subspaces of $\mathbb{C}^n$. The set is nonempty since $\mathbb{C}^n$ is its own (nonzero) subspace which is both $A$- and $B$-invariant ($A\mathbb{C}^n\subset\mathbb{C}^n$ and $B\mathbb{C}^n\subset\mathbb{C}^n$). Hence the set $\mathcal{D}:=\{\dim \mathcal{S}:\mathcal{S}\in\mathcal I\}$ is a nonempty subset of $\{1,\ldots,n\}$. By the well-ordering principle, $\mathcal{D}$ has the least element and hence there is a nonzero $\mathcal{S}\in\mathcal{I}$ of the least dimension.