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Let $A$ and $R$ be row equivalent $m \times n$ matrices. Let the row vectors of $A$ be $a_1, a_2, a_3, \ldots, a_m$ and the row vectors of $R$ be $r_1, r_2, r_3,\ldots, r_m$. Matrices $A, R$ are row equivalent, therefore the $r$ row vectors are obtained from the $a$ row vectors by elementary row operations. This means that every $r$ row vector is a linear combination of the $a$ row vectors. Therefore the row space of matrix $A$ lies in the row space of matrix $R$.

Is the bolded part in the quote saying $R$ is a subset of row space of $A$? How does that show that row space of $A$ is in the row space of $R$?

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It doesn't make any sense to say that "$R$ is a subset of row space of $A$", since $R$ is a matrix and the row space of $A$ is a subspace; they are different objects.

The bolded part implies that every $r$ row vector is in the row space for $A$. That in turn implies that every linear combination of row vectors from $R$ is in the row space for $A$, since the row space of $A$ is a subspace and hence closed under linear combinations. Therefore, the row space of $R$ is in the row space for $A$.

Since we can flip the roles of $R$ and $A$ with no changes to the above, we can also say that the row space of $A$ is in the row space of $R$. Hence, the row space of $R$ and the row space of $A$ are the same.

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    @ Nathan H., row space of A is a subspace of what or how do we know it's a subspace? Thanks. – 2017-01-03
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    The row space of $A$ is a subspace of $\mathbb{R}^n$, if $A$ is $m$ by $n$. It's a subspace because it's closed under vector addition and scalar multiplication. I can provide more details if you need me to. (I would edit my original post, it's too hard to type it out here.) – 2017-01-03
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    @ Nathan H., I see what's up. Cheers. – 2017-01-03