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 $W_1 \subset W_2$.
