Fact: If $a$ and $b$ are vertices of a graph such that there is a path in the graph from $a$ to $b$, then out of all such paths there is a path from $a$ to $b$ of shortest lengths. Additionally, if $v_0,...,v_n$ is a path from $x$ to $a$ of shortest length, then this path traverses $e$ at most once.
Suppose that $a$ and $b$ are vertices of a connected graph $G$ which are endpoints of an edge $e$. Let $x$ be any vertex of $G$.
a) Prove that either there is a path from $x$ to $a$ which does not traverse $e$ or there is a path from $x$ to $b$ which does not traverse $e$.
b) Prove that $G\setminus e$ has either exactly one or exactly two connected components.
I think I have a good conceptual understanding of these two questions, but I am having trouble writing them down formally. For (a), given a vertex $x$ it will have to be on "either side" of the edge $e=\{a,b\}$. So, if there is a path from $x$ to $a$, it will either go through $b$ or not go through $b$.
Case 1: If it does then the path $x,...b,a$ exists. Removing $a$, we get the path $x,....,b$ which does not traverse $e$
Case 2: If it doesn't go through $b$, then $x,...,a$ exists and does not traverse $e$. Q.E.D
As for the second one, I know I need to use what I have just shown that given a vertex of $G\setminus e$, there is a path in $G\setminus e$ from that vertex to one of the endpoints of $e$. If there is a path to both endpoints which does not traverse $e$ then the graph has one connected component. If not, then there are two connected components.
Here is a formal formatting for what you wrote above.
(A)
Let $x \in V(G)$. Since $G$ is connected, there is a shortest $a \to a$ path $P$ in $G$. If $P$ does not traverse $e$, we are done. Assume now $P$ does traverse $e$. By the Fact, $P$ traverses $e$ exactly once, and since $P$ ends on $a$, the last edge of $P$ is $(b,a)=e$. Hence, $P \backslash e$ is a $x \to b$ path in $G$ that does not traverse $e$.
(B)
Now let $H = (V(G), E(G) \backslash e)$ and assume that $H$ has more than one connected component. In such a case, $a$ and $b$ must be in different components of $H$, say $C_a$ and $C_b$. Let's assume for the sake of contradiction that there are more than 2 connected components in $H$. Let $x \in V(G)$ such that $x$ is in the connected component, distinct from $C_a$ and $C_b$. Then, $x,a,b$ are pairwise disconnected in $H$.
However, in $G$, there must have been either $x \to a$ path or $x\to b$ path which do not traverse $e$. WLOG (because of symmetry wrt $a$ and $b$) assume $x \to a$ path exists in $G$ without traversing $e$. Then the same $x \to a$ path exists in $H$ as well, which contradicts that $x$ and $a$ are disconnected in $H$.