Why is $\mathbb{R}^2$ not a subset and /or a subspace of $\mathbb{R}^3$? One thing this suggests--at least to me--is that the x-y plane and $\mathbb{R}^2$ are not necessarily equivalent.  For example, I could define the following: $X = \left\{ \begin{bmatrix} x\\y\\z\end{bmatrix} x,y \in \mathbb{R} \land z = 0\right\}$.  Am I wrong to think, one, that this is a subset of $\mathbb{R}^3$?  As I write this it occurs to me that while scalar multiplication is closed under the above rules, addition doesn't pass the smell test for a subspace... so, OK, it's certainly not a subspace.  I would welcome any insight readers of this query can provide.
The elements of $\Bbb R^2$ are vectors of two coordinates; and the elements of $\Bbb R^3$ are vectors of three coordinates. (One can easily think of those vectors as $2$-tuples and $3$-tuples, for example.)

Assuming mathematics is consistent, $2\neq 3$. Therefore no element of $\Bbb R^2$ is an element of $\Bbb R^3$. It follows that $\Bbb R^2$ is not a subset of $\Bbb R^3$.

And in order to be a subspace, one first has to be a subset. So it's not a subspace either.



What you have defined as $X$ is isomorphic to $\Bbb R^2$, but just as well you could decide that $y$ is $0$, and the identification would still be natural. $X$ is a subset of $\Bbb R^3$ and indeed a subspace, but it is not $\Bbb R^2$ as a set, it is just isomorphic to it in a very obvious way.

While isomorphism is an equivalence relation, and we often think of it almost as identity, it is still not set equality which is a stricter notion.
Assume $m&lt;n.$  Then on the standard basis, $\mathbb{R}^m$ is, in fact a subspace of $\mathbb{R}^n.$
The vector space $\mathbb{R}^n$:
Begin by defining $\mathbb{R}^n$ (and similarly for $\mathbb{R}^m,$ etc.,) as the $n^{th}$ Cartesian product of $\mathbb{R}.$  Doing so requires an ability to distinguish each $\mathbb{R}$ appearing as a Cartesian factor from the others.  For example, in $\mathbb{R}^3$ we distinguish each by calling them the $x,y,z$ axes.  We  nominate each factor using a subscript natural number $i=1 \dots i=n$ so $\mathbb{R}^n=\mathbb{R}_1\times\dots\times\mathbb{R}_n,$ and assume the natural ordering on the indices.  These designations are arbitrary.  We could choose any unique subscript values in any ordering, just so long as we distinguish each Cartesian factor $\mathbb{R}$ from the others.
The standard basis $\left\{\mathfrak{\hat{e}}_i\right\}$ of $\mathbb{R}^n:$
From each $\mathbb{R}_i$ we identify its real number $1\in\mathbb{R}_i$ with the symbol $\mathfrak{\hat{e}}_i\in\mathbb{R}_i.$  The $\mathfrak{\hat{e}}_i$ are called the elements (vectors) of the standard basis of $\mathbb{R}^n.$  The set $\left\{\mathfrak{\hat{e}}_i\right\}$ is linearly independent, and spans $\mathbb{R}^n$.  That is, every element of $\mathbb{R}^n$ is given as a unique linear combination of the standard basis vectors.
$\mathbb{R}^m$ as a subspace of $\mathbb{R}^n$: We construct $\mathbb{R}^m$ and its standard basis $\left\{\mathfrak{\hat{f}}_j\right\}$ using the same method with $j=1 \dots j=m&lt;n.$  The next natural step is to identify $\left\{\mathbb{R}_i=\mathbb{R}_{j=i\le{m}}\right\}$ which gives $\left\{\mathfrak{\hat{f}}_j=\mathfrak{\hat{e}}_{j=i\le{m}}\right\}.$
Therefore, assumming standard definitions and natural identities, $\mathbb{R}^m\subset\mathbb{R}^n.$
Of course, we could use different definitions, leading to different results.  For example, a typical definition is $\mathbb{R}^{n-m}=\mathbb{R}_{i=m+1}\times\dots\times\mathbb{R}_n.$
Other than an appeal to sanity, there is nothing to stop us from identifying $\mathbb{R}^{p}=\mathbb{R}^{n-m},$ where $p=n-m,$ and $\mathbb{R}^{p}$ is constructed as above.  Indeed $\mathbb{R}^{p}$ and $\mathbb{R}^{n-m}$ are isomorphic, so there is an equivalence relation between them.