Kodaira defines a complex analytic family of compact complex manifolds as the data $(E,B,\pi)$, where $E$ and $B$ are complex manifolds, and $\pi$ is a surjective holomorphic submersion such that the preimage $\pi^{-1}(x)$ of any point $x \in B$ is a compact complex submanifold of $E$. (Here complex manifolds are required to be connected.)
A key property of this definition is that this specifies a differentiable fibre bundle. This fact can almost be obtained from the Ehresmann fibration theorem, which might be stated as follows: "Let $X$, $Y$ be differentiable manifolds, and let $f: X \rightarrow Y$ be a proper surjective submersion. Then $f$ is the projection of a differentiable fibre bundle."
In particular, the map $\pi$ defining a family lacks the properness needed to apply Ehresmann's theorem. It is, however, a corollary of the fact that a family gives a fibre bundle that $\pi$ is indeed proper.
Is there a more elementary way of seeing that $\pi$ must be proper?
I think the following argument answers your question. It reminds me a lot of the exercises in the basic topology course I took during my undergrad. Fun times.
Recall that $\pi$ is a submersion, so for any $x$ in $E$ there exists a neighborhood of the form $U \times V$, such that $\pi$ identifies with the projection $U \times V \to V$. Suppose we only know that $E_t := \pi^{-1}(t)$ is compact for any $t$ in $B$, and let's show that $\pi$ is proper.
Let $K \subset B$ be compact and set $L = \pi^{-1}(K)$. Let $(\mathcal U_\alpha)$ be an covering of $L$ by open subsets of $E$. We will show that there exists a finite covering of $E$ by open subsets of $(\mathcal U_\alpha)$, and thus by sets of the original covering.
Take $t \in K$. Find neighborhoods $\mathcal U_1, \ldots \mathcal U_{n(t)}$ that cover $E_t$. For any point $x \in E_t$, find neighborhoods $U_{t,x} \times V_{t,x} \subset \mathcal U_{j(x)}$ for some $j(x)$ (the $\mathcal U_{j(x)}$ being one of the above), such that the map $\pi$ identifies with the projection $U_{t,x} \times V_{t,x} \to V_{t,x}$.
The compactness of $E_t$ gives finitely many such $U_{t,x_\nu} \times V_{t,x_\nu}$ which cover $E_t$ and such that there is a $V_t \subset B$ contained in the image of any $V_{t,x_\nu}$ by $\pi$ (the intersection of the finitely many such $V$ contains $t$).
Now: $K$ is compact, so it is covered by finitely many of those $V_t$. The corresponding finite collection $(U_{t,x_\nu} \times V_{t,x_\nu})$ covers $L$, and consists of subsets of elements of $(\mathcal U_\alpha)$. Thus finitely many elements of $(\mathcal U_\alpha)$ cover $L$.