Dimension of two finite-dimensional subspaces 
Suppose $V$ is a vector space over a field $F$, and $S$, $T$ are two finite-dimensional subspaces of $V$.
Prove that $S \subset T \Rightarrow $ dim($S$) $\leq$ dim($T$)

My solution)
Let $\{s_1,...,s_n\}$ be a basis of $S$.
Take $t \in T \setminus S$.
If $t$ is a linear combination of $\{s_1,...,s_n\}$, then $t \in S$, which is a contradiction.
Thus $t$ is not a linear combination of $\{s_1,...,s_n\}$.
Now $\{s_1,...,s_n,t \}$ is a basis of $S \cup \{{t}\} $.
Therefore, dim($S$) $\leq$ dim($S \cup \{{t}\} $) holds.
Since $t \in T \setminus S$ is arbitrary, dim($S$) $\leq$ dim($S \cup [T  \setminus S ] $) = dim($T$) holds.
Is it correct?

The last line is not necessary.   Note:  $$dimT \ge dim (S\cup span\{t\})= dimS+1\gt dimS $$.  The first inequality because there are $dimS+1$ independent vectors in  $T $...
So, if $\exists t \notin S $ then $dimS\lt dim T $, otherwise  $S=T $ so $dimS=dimT $...
