On page 292 of Hatcher's algebraic topology, he writes that "Since the columns of a matrix in $O(n)$ are unit vectors, $O(n)$ can also be regarded as a subspace of the product of $n$ copies of $S^{n-1}$. It is a closed subspace since the condition that columns be orthogonal are defined by polynomial equations."
I don't understand why we can regard $O(n)$ as a subspace of the product of $n$ copies of $S^{n-1}$ and I don't understand why the columns being orthogonal are defined by polynomial equations and then how that tells us that the subspace is closed.
The sphere $S^{n-1}\subset \Bbb R^n$ is the set of vectors of length $1$ in $\Bbb R^n$, hence $X:=S^{n-1}\times\ldots \times S^{n-1}$ (with $n$ factors) is the set of $n$-tuples of $n$-vectors of length $1$.
Since an orthogonal matrix (viewed as their list of column vectors) is an $n$-tuple of $n$-vectors of length $1$, it is an element of $X$, i.e., $O(n)\subseteq X$.
The conditions that distinguish an element of $O(n)$ from a general element of $x$ are of the form $a_1b_1+a_2b_2+\ldots+a_nb_n=0$. Since the left hand side is a polynomial (of degree $2$) in $a_1,\ldots, a_n,b_1\ldots, b_n$, this condition (as well as all its cousins) is a "closed" condition: The inverse image of the closed set $\{0\}$ under the polynomial (a continuous function!) is closed. Combining the finitely many orthogonality conditions means to intersect finitely many closed sets - and results in a closed set.