First take note, our professor likes to change terms from our book. I have the following:
Assume we have a connected graph $G$ where all of the vertices have even degree. If we assume it does have a “cut-vertex” / bridge. Let $u$ and $v$ be the two vertices that connects to form our edge $e$. So know we will have two components (call one $H$ and $I$) consisting of $G - e$. Originally all of the vertices of $G$ had a even degree. But since we have removed one edge from at least a vertex in each component, that means we have an odd vertex in each component ($\mathrm{even}\,– 1 = \mathrm{odd}$). But by the corollary (1.5 in book) to the First Theorem of Graph Theory, every graph must have a even number of odd vertices, which neither component of $G$ has.
Is this sufficient? I have read other proofs on here and not sure if mine will work.