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This is a part of the proof (see title):

Assume a graph $G$ is connected and each degree is even. Let $W=(v_0,v_1,\dots,v_n)$ be the longest possible walk$^{(*)}$ where each edge occurs at most once. We know that $W$ is a closed walk.

Assume now that $W$ is not an Euler cycle. This means that there is an edge that is not crossed in $W$, say $f$. We know that $G$ is connected, so there is a path $P=(w_0,w_1,\dots,w_m)$ such that $f=\{w_0,w_1\}$ and $w_m\in W$. We choose $P$ as small as possible. This means that the edges in $P$ don't occur in $W$ (otherwise we could shorten $P$).

Why is it true that we can shorten the length of $P$ is an edge of $W$ is in $P$? We know that $W$ is the longest possible walk, such that each edge occurs at most once. How does the occurrence of an edge of $W$ open up a possibility to shorten $P$? I tried reasoning this, by considering a pair of points that are connected by this edge in $W$, say $\{w_i,w_{i+1}\}$. I guess $(w_i,\dots,w_m)$ could be shorter now. But how?

$^{(*)}$ A walk is a sequence $(w_0,w_1,\dots,w_k)$, such that $w_i$ and $w_{i+1}$ are connected for each $i\in\{1,\dots,k\}$.

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    If, say, $(w_i,w_{i+1}) \in W$, then $i < m$, and you simply cut $P$ off at $w_i$ to obtain a shorter path from $(w_0,w_1)$ to $W$. – 2017-02-25
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    @MeesdeVries Ohh, I had forgotten that this $w_m$ was just a random point in $W$. Thank you! – 2017-02-25

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