Let the sequence be $ a_{1}, a_{2}, a_{3},\ldots$. Define a new sequence $ b_{k}=\lfloor a_{k}/ 10\rfloor$. We note that $ a_{k}+10 > a_{k+1}$, so $ b_{k}+1\geq b_{k+1}$. Thus $ \{b_{i}\}$ only increases by one at a time. Furthermore, the sequence $ \{b_{i}\}$ is unbounded since $ \{a_{i}\}$ is unbounded. Combined, we see that $ \{b_{i}\}$ assumes every integer greater than or equal to $ b_{1}$. Suppose $ b_{1}$ has $ m$ digits with decimal representation $ d_{0}d_{1}d_{2}\ldots d_{m}$. Now define $ d^{\prime}_{k}$ to be the least odd number greater than or equal to $ d_{k}$. It's clear that $ 0\leq d^{\prime}_{k}< 10$ still holds. So $ d^{\prime}_{0}d^{\prime}_{1}d^{\prime}_{2}\ldots d^{\prime}_{m}$ is a decimal representation of a number. Call this number $ b^{\prime}$. Notice that $ b^{\prime}$ has all odd digits, and $ b^{\prime}> b_{1}$. This means that at least one term in the sequence $ \{b_{i}\}$ is equal to $ b^{\prime}$, and thus at least one term in the sequence has all odd digits. Let $ k$ be such that $ b_{k}$ has only odd digits. We prove that $ a_{k}$ is even or $ a_{k+1}$ is even. Suppose $ a_{k}$ is odd. Then $ a_{k}$ ends in an odd digit; call it $ d$. We have $ a_{k}= d+10\cdot b_{k}$. Note that this number has all odd digits. So $ a_{k+1}$ will be $ a_{k}$, an odd number, plus a digit of $ a_{k}$, also odd. Thus $ a_{k+1}$ is even. This finishes the proof.

Moreover, we can prove that we can find as long as we want sequences of odd integers.