Showing that $\dim( H \cap K) \leq \dim(H)$? In the text "Linear Algebra and it's Applications" by David C Lay, Stevan R Lay and Judi J McDonald I'm inquiring if my solution for the following problem is correct ?
$\textbf{Problem}$
Prove that,
$$\dim( H \cap K) \leq \dim(H)$$
$\text{Solution}$
To establish the inequality introduce respectively,
$$H=\left[\begin{array}{cccc}h_{11} & h_{12} & \cdots & h_{1 n} \\ a_{21} & h_{22} & \cdots & h_{2 n} \\ \vdots & \vdots & \vdots & \vdots \\ h_{m 1} & h_{m 2} & \cdots & h_{m n}\end{array}\right], K=\left[\begin{array}{cccc}k_{11} & a_{12} & \cdots & k_{1 n} \\ k_{21} & k_{22} & \cdots & k_{2 n} \\ \vdots & \vdots & \vdots & \vdots \\ k_{m 1} & k_{m 2} & \cdots & k_{m n}\end{array}\right]$$
Putting the icing on the cake, and lastly via Theorem (11) we see that,
$$\dim( H \cap K) \leq \dim(H) = \dim \Bigg( \left[\begin{array}{cccc}h_{11} & h_{12} & \cdots & h_{1 n} \\ a_{21} & h_{22} & \cdots & h_{2 n} \\ \vdots & \vdots & \vdots & \vdots \\ h_{m 1} & h_{m 2} & \cdots & h_{m n}\end{array}\right] \bigcap \left[\begin{array}{cccc}k_{11} & a_{12} & \cdots & k_{1 n} \\ k_{21} & k_{22} & \cdots & k_{2 n} \\ \vdots & \vdots & \vdots & \vdots \\ k_{m 1} & k_{m 2} & \cdots & k_{m n}\end{array}\right] \Bigg) \leq \dim \Bigg(\left[\begin{array}{cccc}h_{11} & h_{12} & \cdots & h_{1 n} \\ a_{21} & h_{22} & \cdots & h_{2 n} \\ \vdots & \vdots & \vdots & \vdots \\ h_{m 1} & h_{m 2} & \cdots & h_{m n}\end{array}\right] \Bigg)$$
$$  \, \, \, \, \, \, \, \, \, \,    = \dim \Big( \sum h_{i}a_{i} \, \bigcap \, \, \sum k_{i}a_{i} \Big) \leq \dim \Big( \sum h_{i}a_{i} \Big)$$
$$ \! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \! =  p \leq \dim \Big( \sum h_{i}a_{i} \Big). $$
QED.

I assume that $H$ and $K$ are subspaces of an ambient vector space $V$. By the Theorem you linked [Theorem (11)], since $H\cap K$ is a subspace of $H$, it follows that $\dim H\cap K\le \dim H$.
You don't need to appeal to bases at all once you know the theorem you linked.
