We will prove by induction that $\text{odd}(N) \cdot \text{even}(N) < N$ for all positive integers $N$ with $2k$ digits.

If $N = 10a + b$, $a, b \in \{0, 1, 2, \dots, 9\}$, $a \neq 0$, $N = 10a + b > a \cdot b + b \ge a \cdot b = \text{even}(N) \cdot \text{odd}(N)$.

Now suppose that $N$ has $2k > 2$ digits and that the claim is true for all numbers with $2k-2$ digits. Let $c$ and $d$ be the two leftmost digits of $N$, so that $N = c \cdot 10^{2k-1} + d \cdot 10^{2k-2} + N_0$, $N_0$ with $2k-2$ digits. Then $\text{odd}(N) = d \cdot 10^{k-1} + \text{odd}(N_0)$ and $\text{even}(N) = c \cdot 10^{k-1} + \text{even}(N_0)$. So we need to prove that
$$\begin{align*} c \cdot 10^{2k-1} + d \cdot 10^{2k-2} + N_0 &> (c \cdot 10^{k-1} + \text{even}(N_0)) \cdot (d \cdot 10^{k-1} + \text{odd}(N_0)) \\ \iff c \cdot 10^{2k-1} + d \cdot 10^{2k-2} + N_0 &> cd \cdot 10^{2k-2} + d \cdot 10^{k-1} \cdot \text{even}(N_0) + c \cdot 10^{k-1} \cdot \text{odd}(N_0) + \text{odd}(N_0) \cdot \text{even}(N_0) \end{align*}$$
But this is true, since both $\text{odd}(N_0)$ and $\text{even}(N_0)$ are less than $10^{k-1}$ and thus
$$\begin{align*} c \cdot 10^{2k-1} &\ge c(d+1) \cdot 10^{2k-2} > cd \cdot 10^{2k-2} + c \cdot 10^{k-1} \cdot \text{odd}(N_0) \\ d \cdot 10^{2k-2} &> d \cdot 10^{k-1} \cdot \text{even}(N_0) \\ N_0 &> \text{odd}(N_0) \cdot \text{even}(N_0) \end{align*}$$