On the Dimensionality of Space: An Elementary Analysis The below theorem I am to prove. Perhaps you have a critique...

Theorem 2.4 Let $W_1$ and $W_2$ be two subspaces of a vector space $V$. Then $\dim(W_1 \cap W_2)=\dim(W_1)$ if and only if $W_1 \subset W_2$.

My—perhaps lazy—attempt to prove it... I say lazy in the sense that I don't show $\subseteq$ then $\supseteq$.
Let $W_1$ and $W_2$ be finite subspaces of the vector space $V$. Let
\begin{eqnarray}
\{\alpha_1,\alpha_2,\dots,\alpha_l\}
\end{eqnarray}
be a basis for $W_1 \cap W_2$. Because $W_1 \cap W_2 \subset W_1$, then we have
\begin{eqnarray}
\{\alpha_1,\alpha_2,\dots,\alpha_l,\beta_1,\beta_2,\dots,\beta_m\}
\end{eqnarray}
as a basis for $W_1$, and similarly we have
\begin{eqnarray}
\{\alpha_1,\alpha_2,\dots,\alpha_l,\gamma_1,\gamma_2,\dots,\gamma_n\}
\end{eqnarray}
as a basis for $W_2$, so $W_1+W_2$ is spanned by the linearly independent set of vectors
\begin{eqnarray}
\alpha_1,\dots,\alpha_l,\beta_1,\dots,\beta_m,\gamma_1,\dots,\gamma_n
\end{eqnarray}
and has dimension $l+m+n$. Thus, from (14) and (15) we have that $\dim W_1 = l+m$ and $\dim W_2 = l+n$, so
\begin{eqnarray}
\dim W_1 + \dim W_2 & = & (l+m)+(l+n) \\
&=&l+(l+m+n) \\
&=&\dim(W_1 \cap W_2) + \dim(W_1+W_2).
\end{eqnarray}
This being the case, then the second statement in Theorem 2.4 implies that $\dim W_2 = \dim (W_1+W_2)$, or $l+n=l+m+n$. This means that $m=0$, which implies that the basis for $W_1$ has the form
\begin{eqnarray}
\{\alpha_1,\dots,\alpha_l\},
\end{eqnarray}
so since $W_2$ has the form
\begin{eqnarray}
\{\alpha_1,\dots,\alpha_l,\gamma_1,\dots,\gamma_n\},
\end{eqnarray}
then it is clear that this can only be the case when $W_1 \subset W_2$.

Depending on how well is one acquainted with these things I'd prove this much shorter and using basic set theory:
$$W_1\subset W_2\iff W_1\cap W_2=W_1$$
and thus for finite dimensional vector subspaces:
$$W_1\subset W_2\iff \dim(W_1\cap W_2)=\dim W_1$$
