Part 1 - Eulerian Circuit:

($\Rightarrow$) Necessity: Suppose an Eulerian circuit exists. Consider any vertex $v$. Each time the circuit passes through $v$, it enters via one edge and leaves via another edge (except at the start/end where it both starts and ends). Since the circuit uses each edge exactly once, the number of times it visits $v$ equals $\deg(v)/2$. Thus $\deg(v)$ must be even.

($\Leftarrow$) Sufficiency: Assume all vertices have even degree. We construct an Eulerian circuit by algorithm:

Start at any vertex $v_0$ and follow edges arbitrarily, never repeating an edge, until stuck. Call this walk $W$.

Claim: When stuck, we must be back at $v_0$.

Proof: If stuck at vertex $u \neq v_0$, we entered $u$ the same number of times we left it, plus one final entry. This means we used an odd number of edges at $u$. But $\deg(u)$ is even, so unused edges remain—contradiction.

If $W$ uses all edges, we're done. Otherwise, since $G$ is connected, some vertex $v$ on $W$ has an unused incident edge. Starting from $v$, construct another walk $W'$ using only unused edges (which form a graph with all even degrees at remaining vertices). Continue recursively merging walks until all edges are used.

Part 2 - Eulerian Path:

($\Rightarrow$) If an Eulerian path exists from $u$ to $v$ with $u \neq v$, then $u$ and $v$ have odd degree (starting and ending use odd numbers of edges at these vertices), while all other vertices have even degree (enter and leave equal numbers of times).

If $u = v$, this is an Eulerian circuit, so all degrees are even.

($\Leftarrow$) If exactly two vertices $u,v$ have odd degree, add edge $\{u,v\}$ to create graph $G'$. Now all degrees in $G'$ are even, so $G'$ has an Eulerian circuit by Part 1. Remove edge $\{u,v\}$ from this circuit to obtain an Eulerian path in $G$ from $u$ to $v$.