Proof Verification: A subspace $W$ of an n-dimensional vector space $V$ is a finite dimensional vector space of dimension $m\leq n$.

Question

Proof:

Suppose $W$ is a subspace of vector space $V$ then by definition of a subspace, it is a nonempty subset of $V$ and is itself a vector space. Note that the dimension of a vector space $V$ is the number of elements in its basis. By the Steinitz Replacement Theorem, $V$ has at most $n$ linearly independent elements for any subset $A\subseteq V$. Since $W$ is nonempty and $W\subseteq V$, then any basis $X\subseteq W$ has at most $n$ linearly independent elements. Thus, if $X$ is a basis then it has at most $n$ elements which implies $dimW=m\leq n$

Answer 1

Your proof is OK, apart from some awkward wording. In particular, I don't like the sentence

$V$ has at most $n$ linearly independent elements for any subset $A\subseteq V$

I think what you are trying to say is

Any linearly independent set $A\subseteq V$ has at most $n$ elements.

Also, a short word about how you know that if $X$ is linearly independent in $W$, then it is also linearly independent in $V$ would be nice.

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So the proof would be reworded as

By the Steinitz Replacement Theorem, any linearly independent set $A\subseteq V$ has at most $n$ elements.

Let $X$ be a basis for $W$. Then, $X$ is a linearly independent set in $W$. Let $m$ be the number of elements of $X$ (therefore, $m=\dim W$). Therefore, $X$ is also a linearly independent subset of $V$ (just a short description how you know that would be nice here), and therefore, $X$ has at most $n$ elements, or in other words, $m\leq n$.