A finite graph with exactly two vertices with odd degree must have a path joining them.

Assume $G$ is a finite graph with exactly $2$ vertices of odd degree, call them $u$ and $v$. If $G$ is connected, then we are finished. Thus we can assume $G$ is disconnected. Let $J$ be the connected component of $G$ such that $u\in J$. Since $J$ is a graph, it must contain an even number of vertices of odd degree since

$$\sum_{w\in V(J)} \text{deg}(w)=2|E(J)|.$$

Thus there is at least one more vertice of odd degree in $J$. Since $G$ contains exactly two vertices of odd degree, we must have $v\in J$. This implies there is a path from $u$ to $v$.

"Let $G$ be a disconnected finite graph...." Why? The proof should start with "If $G$ is connected, then of course there's a path between any two vertices, so assume $G$ is a disconnected finite graph...."

"...with exactly two vertices ... with odd degrees..." Why? You are trying to prove something about graphs with at least two vertices of odd degree, not exactly two vertices of odd degree. And why two components? I think it should go, "...with at least two vertices, $u$ and $v$, of odd degree."

"Then either of the two possibilities given below must hold such that [the] number of edges in $G$ is a positive integer." This is not English. The clause beginning "such that" doesn't make sense. What are you trying to say?

You give no reason why one of the two possibilities must hold. You write ungrammatical sentences (the two sentences with the word, "contradiction", are not grammatically correct).

On top of this, there is Didier's objection that what you are trying to prove is false. Maybe you really do want the hypothesis to state "exactly" instead of "at least". Or maybe you want the conclusion to be "a path joining some pair of vertices of odd degree."