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?
 A: 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 $...
A: Note that $\dim(S\cup\{t\})$ doesn't make sense, because $S\cup\{t\}$ is generally not a subspace (it definitely isn't if $t\in T\setminus S$, by the way).
You could try and salvage it by using $S+\operatorname{span}\{t\}$, but you cannot claim $\dim(S+\operatorname{span}\{t\})\le\dim T$, because this uses what you're trying to prove.


Proposition. The dimension of a (finitely generated) vector space is the maximal size of a linearly independent set.

Proof. Suppose $\{v_1,\dots,v_m\}$ is a linearly independent set with maximal size in the vector space $V$. If it is not a basis, then there exists $v$ which is not a linear combination of $\{v_1,\dots,v_m\}$. In this case, $\{v_1,\dots,v_m,v\}$ is a linearly independent set, contradicting maximality. QED
With this proposition, you have the proof for your statement: assuming $\dim S>\dim T$ would provide a linearly independent set in $S$ with size larger than the dimension of $T$: contradiction.
A: Let $\{v_1, \ldots, v_k\}$ be a basis for $S$.
$S$ is a subspace of $T$, so the linearly independent set $\{v_1, \ldots, v_k\}$ can be extended to a basis $\{v_1, \ldots, v_k, v_{k+1}, \ldots, v_n\}$ for $T$. Obviously $k \le n$.
By definition of dimension (which is the cardinality of any basis), we have $\dim S = k$ and $\dim T = n$. Therefore, $\dim S \le \dim T$.
