Prove that for every positive integer $k$, every connected graph with exactly $2k$ vertices of odd degree and arbitrarily many vertices of even degree decomposes into $k$ trails. Does this remain true without the connectedness hypothesis?
So I have so far attempted to prove this with induction. The base case would be $k=1$ for which you would get a graph $K_{1,1}$ which has $2$ odd vertices so it fits the graph description and has only a single trail so it has $k$ trails. I just fail to see how this continues going up with potentially infinite even vertices attached.
You're on the right track. As you suspect, connectedness is not necessary. In fact, the induction proof is easier if you drop it as an assumption.
Your base case is not as robust as it needs to be. You don't just have to deal with a single graph with $2$ odd vertices, you have to deal with any such graph. However, this is not hard to do. If you have a graph with $2$ odd vertices, then connecting those vertices will result in an Eulerian graph. This graph can be covered in a cycle, and removing that added edge gives us a trail connecting the two odd vertices.
Here is the induction step.
Assume every graph with $2k$ odd vertices can be decomposed into the union of $k$ trails with $k\ge1$. Let $G$ be a graph with $2k+2$ odd vertices. Therefore, we may choose some odd vertex $u$. Since every component of a graph has an even number of odd vertices, we may choose an odd vertex $v$ that is in the same component as $u$. Since $u$ and $v$ are in the same component, we may choose $T$ to be a $(u,v)$-trail of $G$. Now, consider $G-T$. It is easily seen that the degrees of $u$ and $v$ must now be even and the degree of every other vertex of $G-T$ is the same parity as that vertex of $G$. In other words, $G-T$ has exactly $2k$ odd vertices. By the induction hypothesis, we may decompose $G-T$ into $k$ trails, so $G$ must be the union of $k+1$ trails.
(The place where the proof becomes easier if you work with all graphs and not just connected graphs is that with our naive approach there is no way to be certain that $G-T$ is connected.)