The problem you're asking about is Exercise 3.3.16 on p. 147 of the second edition of West's Introduction to Graph Theory. The "even order" assumption is of course redundant when $k$ is odd. The proposition is trivial if $k=1$ ($G=K_2$) or $k=2$ ($G$ is an even cycle). The case $k=3$ does not seem trivial. Leafing through the preceding pages of text, I found the case $k=3$ stated and proved as Corollary 3.3.8 on p. 139. If I had to work this exercise, the first thing I'd do is study that proof carefully. Then I try to generalize it, first for $k=4$, then for general $k$. Also, hoping it's not too much of a spoiler, I might look for a hint in Appendix C: Hints for Selected Exercises. The hint for this one is on p. 511, and it just says "Generalize the argument of Corollary 3.3.8."
But maybe you've already read that hint, and you need help in figuring out how to generalize the proof of 3.3.8? In that's where you're stuck, you should have said so.
Assume for sake of contradiction that $G$ does not have a perfect matching. By Tutte's theorem, there exists a set $S$ such that $G - S$ has more odd components than there are vertices in $S$. Let $A_1, \ldots, A_t$ be the odd components of $G - S$. Notice that $t + |S|$ is even, since $G$ has an even number of vertices, and therefore $|S| < t$ implies $|S| + 2 \leq t$.
Notice that it is impossible that there are exactly $r-1$ edges between $S$ and one of the $A_i$. This is because the degree sum of the subgraph induced by $A_i$ would then be $r|A_i| - (r - 1)$ which is an odd number, and by the famous handshake lemma we can't have an odd degree sum.
The number of edges between $S$ and $A_i$ is at least $r-2$, since $G$ is $r-2$ edge connected. At most $r-1$ of the $A_i$ have $r-2$ edges to $S$. As discussed in the previous paragraph, the other $(t - (r-1))$ components must have at least $r$ edges to $S$.
So how many edges are there leaving $S$? Well on the one hand this is at most $r |S| \leq r(t - 2) = rt - 2r$, since the vertices each have degree $r$. On the other hand, this is at least (counting by how many edges are between $S$ and each $A_i$) the following: $(r-2)(r-1) + r(t - (r-1)) = rt- 2r + 2$. Thus we have a contradiction which proves the result.
Best Answer
Lemma. In a $k$-regular graph $G$, if $S \subseteq V(G)$ with $|S|$ odd, there cannot be exactly $k-1$ edges between $S$ and $V(G)-S$.
Proof. Suppose there are $k-1$ such edges. Then $\sum_{v \in S}\deg(s) = k|S|$ counts each edge between two vertices in $S$ twice, and each edge from $S$ to $V(G)-S$ once, so there are $\frac{k|S|-(k-1)}{2}$ edges between two vertices in $S$. But this simplifies to $k \cdot \frac{|S|-1}{2} + \frac12$, which is not an integer. $\square$
Now we are ready to apply Tutte's theorem. Let $U$ be any nonempty subset of $V(G)$. Then there are at most $k|U|$ edges out of $U$. If $C$ is an odd component of $G-U$, it must receive at least $k$ of those edges: there cannot be $k-2$ or fewer by the connectivity condition, and there cannot be $k-1$ by the lemma. Therefore there can be at most $\frac{k|U|}{k}$ odd components.
If $U = \varnothing$, then the connectivity condition does not apply, because there might just be one odd component $C$ which is the whole graph. This is where we need the assumption that $|V(G)|$ is even.