If $\dim(M+N) = \dim(M)$, then since $M \subset M+N$, we have $M = M+N \supset N$ (contradicting $N \not\subset M$)
Therefore $\dim(M+N) > \dim(M) = n-1,$ which along with $\dim(M+N) \leq n$ implies $\dim(M+N) = n$.
Now finish by the dimension theorem.
I've taught college linear algebra to qualified high school students, but a 14-year old would typically be a high school freshman. I'd therefore try to frame the notions in concrete vector space terms, e.g. working with vector space $V$ consisting of tuples $(x_1,x_2,\ldots,x_n)$ of real numbers.
It is a nice property of finite dimensional vector spaces that their subspaces are again finite dimensional. While one might be happy to "verify with examples" your statement (2), even to take this approach to the dimension of a sum $W+U$ of subspaces of $V$ will require familiarity with how dimension of a vector (sub)space is defined:
The dimension of a vector space $V$ is $n$ iff $V$ has a basis set containing exactly $n$ vectors.
In other words one certainly needs to define a basis for a vector space, and to know pertinent facts about subspaces such as how a subspace is defined and:
- The intersection $W\cap U$ of two subspaces of $V$ is also a subspace of V.
- A basis for any subspace $W$ of finite dimensional $V$ can be extended to a basis for $V$.
- Every basis for subspace $W$ has the same size (so the dimension of $W$ is well-defined in not depending on the choice of a basis).
Let me dwell briefly on another crucial idea, the sum of two subspaces $W+U$. There are two competing (but equivalent) ways to define this:
(a) The sum $W+U$ is the intersection of all subspaces of $V$ that contain both $W$ and $U$.
(b) The sum $W+U$ is the collection of all vectors in $V$ that can be expressed as $w+u$ for some $w\in W$ and $u\in U$.
In either case there is a bit work required to establish that $W+U$ is also a subspace of $V$. It would then be possible to present the chain of ideas I laid out in my Comment.
The intersection $W\cap U$ is a subspace and has a finite basis $\{v_1,\ldots,v_k\}$. Thus $\dim(W\cap U) = k$, the size $k\ge 0$ of its basis.
This basis of $W\cap U$ can be extended to a basis $$\{v_1,\ldots,v_k,w_1,\ldots,w_\ell\}$$ of subspace $W$, and similarly extended to a basis $$\{v_1,\ldots,v_k,u_1,\ldots,u_m\}$$ of subspace $U$.
Finally, and the hardest part, one can show the union of those two basis sets $$\{v_1,\ldots,v_k,w_1,\ldots,w_\ell,u_1,\ldots,u_m\}$$ is a basis for $W+U$. How one manages this depends on the pedagogical choice made to define $W+U$ discussed above.
In any case we can now compute the dimension of $W+U$ to be $k+\ell+m$, so that:
$$\dim (W+U) = (k+\ell) + (k+m) - k = \dim(W) + \dim(U) - \dim(W\cap U) $$
Best Answer
My previous attempt is not in the right direction. Following is the proof I have just figured out.
The inequality is equivalent to $\dim S+\dim T-\dim(\overrightarrow{S}+\overrightarrow{T})<\dim X$ where the left side is equal to $\dim(\overrightarrow{S}\cap\overrightarrow{T})$ due to the Grassmann relation. If we can show $\dim(\overrightarrow{S}+\overrightarrow{T})<\dim X$, then $\dim(\overrightarrow{S}\cap\overrightarrow{T})\le\dim(\overrightarrow{S}+\overrightarrow{T})<\dim X$, which establishes the proposition.
To this end, we need to make use of two facts:
Fact 1. is not hard to prove. A proof of Fact 2. is here.
From fact 1., we can infer that $\overrightarrow{S}+\overrightarrow{T}$ is independent of $K\vec{ab}$. To show this, if $(\overrightarrow{S}+\overrightarrow{T})\cap K\vec{ab}$ contains an element $c\vec{ab}\ne 0$, then $c\ne0$, so we have $\vec{ab}\in (\overrightarrow{S}+\overrightarrow{T})$, which is forbidden by Fact 1., because $S\cap T=\emptyset$. Taking dimension on both sides in Fact 2., we have $\dim\overrightarrow{\langle S\cup T\rangle}=\dim\bigl((\overrightarrow{S}+\overrightarrow{T})+K\vec{ab}\bigr)$. The vector spaces in the parentheses on the right side are independent as we have just shown, so we have $\dim\overrightarrow{\langle S\cup T\rangle}=\dim(\overrightarrow{S}+\overrightarrow{T})+\dim K\vec{ab}=\dim(\overrightarrow{S}+\overrightarrow{T})+1$. $\dim\overrightarrow{\langle S\cup T\rangle}\le\dim\vec X$ in general, so we finally get $\dim(\overrightarrow{S}+\overrightarrow{T})<\dim\vec X$.