I found myself working on this same problem (for homework), and I think I've written a fairly detailed solution. So I will post it here, in case it is helpful to anyone else.
Let $\mathfrak{g}$ be a 1-dimensional Lie algebra, and let $\{E_1\}$ be a basis for $\mathfrak{g}$. Then for any two vector fields $X,Y\in\mathfrak{g}$, we have $X=aE_1$ and $Y=bE_1$, for some $a,b\in\mathbb{R}$. Thus,
$$[X,Y]=[aE_1,bE_1]=ab[E_1,E_1]=0$$
for all $X,Y\in\mathfrak{g}$. Therefore, the only 1-dimensional Lie algebra is the trivial one. The map
$$\varphi:\mathfrak{g}\rightarrow\mathfrak{gl}(2,\mathbb{R})$$
$$\varphi:aE_1\mapsto
\left(\begin{array}{ll}
a&0\\
0&0
\end{array}\right)$$
is a Lie algebra homomorphism, since
$$\varphi([aE_1,bE_1])=\varphi(0)=\left(\begin{array}{ll}
0&0\\
0&0
\end{array}\right)\mbox{, and}$$
$$[\varphi(aE_1),\varphi(bE_1)]=\left(\begin{array}{ll}
a&0\\
0&0
\end{array}\right)\left(\begin{array}{ll}
b&0\\
0&0
\end{array}\right)-\left(\begin{array}{ll}
b&0\\
0&0
\end{array}\right)\left(\begin{array}{ll}
a&0\\
0&0
\end{array}\right)$$
$$=\left(\begin{array}{ll}
0&0\\
0&0
\end{array}\right).$$
Thus, $\mathfrak{g}$ is isomorphic to the (abelian) Lie subalgebra
$$\varphi(\mathfrak{g})=\left\{\left(\begin{array}{ll}
a&0\\
0&0
\end{array}\right)\in\mathfrak{gl}(2,\mathbb{R}):a\in\mathbb{R}\right\}\subset\mathfrak{gl}(2,\mathbb{R}).$$
Now let $\mathfrak{h}$ be a 2-dimensional Lie algebra, and let $\{E_1,E_2\}$ be a basis for $\mathfrak{h}$. Then for any two vector fields $X,Y\in\mathfrak{h}$, we have $X=aE_1+bE_2$ and $Y=cE_1+dE_2$, for some $a,b,c,d\in\mathbb{R}$. Thus,
$$\begin{array}{ll}
[X,Y]&=[aE_1+bE_2,cE_1+dE_2]\\
&=a[E_1,cE_1+dE_2]+b[E_2,cE_1+dE_2]\\
&=ac[E_1,E_1]+ad[E_1,E_2]+bc[E_2,E_1]+bd[E_2,E_2]\\
&=(ad-bc)[E_1,E_2].
\end{array}$$
If $[E_1,E_2]=0$, then we have the trivial 2-dimensional Lie algebra. The map
$$\varphi:\mathfrak{h}\rightarrow\mathfrak{gl}(2,\mathbb{R})$$
$$\varphi:aE_1+bE_2\mapsto
\left(\begin{array}{ll}
a&0\\
0&b
\end{array}\right)$$
is a Lie algebra homomorphism, since
$$\varphi([aE_1+bE_2,cE_1+dE_2])=\varphi(0)=\left(\begin{array}{ll}
0&0\\
0&0
\end{array}\right)\mbox{, and}$$
$$[\varphi(aE_1+bE_2),\varphi(cE_1+dE_2)]=\left(\begin{array}{ll}
a&0\\
0&b
\end{array}\right)\left(\begin{array}{ll}
c&0\\
0&d
\end{array}\right)-\left(\begin{array}{ll}
c&0\\
0&d
\end{array}\right)\left(\begin{array}{ll}
a&0\\
0&b
\end{array}\right)$$
$$=\left(\begin{array}{ll}
0&0\\
0&0
\end{array}\right).$$Furthermore, this map is faithful (injective). Thus, $\mathfrak{h}$ is isomorphic to the (abelian) Lie subalgebra
$$\varphi(\mathfrak{h})=\left\{\left(\begin{array}{ll}
a&0\\
0&b
\end{array}\right)\in\mathfrak{gl}(2,\mathbb{R}):a,b\in\mathbb{R}\right\}\subset\mathfrak{gl}(2,\mathbb{R}).$$
If $[E_1,E_2]\neq0$, then set $E_3=[E_1,E_2]$. Then for all $X,Y\in\mathfrak{h}$ we have $[X,Y]=\lambda E_3$ for some $\lambda\in\mathbb{R}$. In particular, for any $E_4\in\mathfrak{g}$ such that $E_4$ and $E_3$ are linearly independent, we have $[E_4,E_3]=\lambda_0 E_3$. Replacing $E_4$ with $1/\lambda_0 E_4$, we now have a basis $\{E_4, E_3\}$ for $\mathfrak{g}$ such that $[E_4, E_3]=E_3$. The map
$$\varphi:\mathfrak{h}\rightarrow\mathfrak{gl}(2,\mathbb{R})$$
$$\varphi:aE_4+bE_3\mapsto
\left(\begin{array}{ll}
a&b\\
0&0
\end{array}\right)$$
is a Lie algebra homomorphism, since
$$\varphi([aE_4+bE_3,cE_4+dE_3])=\varphi((ad-bc)E_3)=\left(\begin{array}{ll}
0&ad-bc\\
0&0
\end{array}\right)\mbox{, and}$$
$$[\varphi(aE_4+bE_3),\varphi(cE_4+dE_3)]=\left(\begin{array}{ll}
a&b\\
0&0
\end{array}\right)\left(\begin{array}{ll}
c&d\\
0&0
\end{array}\right)-\left(\begin{array}{ll}
c&d\\
0&0
\end{array}\right)\left(\begin{array}{ll}
a&b\\
0&0
\end{array}\right)$$
$$=\left(\begin{array}{ll}
0&ad-bc\\
0&0
\end{array}\right).$$Furthermore, this map is faithful (injective). Thus, $\mathfrak{h}$ is isomorphic to the (non-abelian) Lie subalgebra
$$\varphi(\mathfrak{h})=\left\{\left(\begin{array}{ll}
a&b\\
0&0
\end{array}\right)\in\mathfrak{gl}(2,\mathbb{R}):a,b\in\mathbb{R}\right\}\subset\mathfrak{gl}(2,\mathbb{R}).$$
I claim that a morphism $f:\mathfrak{g}\rightarrow\mathfrak{h}$ is monic iff it is injective.
Of course, in any concrete category, injective implies monic, so we'll focus on the more fun direction.
So, assume $f$ is monic and let $\mathfrak{k}$ denote the kernel of $f$. Consider the two functions $g_1,g_2:\mathfrak{k}\rightarrow \mathfrak{g}$ where $g_1$ is the inclusion and $g_2$ is the zero map. Then $f\circ g_1 = f\circ g_2$ since both are the $0$-map. As $f$ is monic, we conclude that $g_1 = g_2$, that is, the that identity map of $\mathfrak{k}$ is the zero map. Thus, $\mathfrak{k} = \{0\}$, so $f$ is injective.
Best Answer
Abstract classification
A Lie algebra $\mathfrak{g}$ is a vector space (over some field $k$) with a bilinear pairing $[\cdot, \cdot]\colon \mathfrak{g}\times\mathfrak{g}\to \mathfrak{g}$ called a bracket, which satisfies
If the dimension of $\mathfrak{g}$ is $n$, then as a vector space this is just $k^n$. So the question is really what brackets can we define on a $1$ and $2$ dimensional vector space?
n=1. For $x,y\in\mathfrak{g}$ we have $cx=y$ for some $c\in k$ and hence $[x,y]=[x,cx]=c[x,x]=0$. In particular, the only bracket one can put on a 1 dimensional vector space is the trivial one: $[x,y]=0$ for all $x,y\in \mathfrak{g}$
n=2. Let $e_1,e_2$ be a basis for this vector space. Carrying out the same computation one finds $$[ae_1+be_2,ce_1+de_2]=(ad-bc)[e_1,e_2].$$ If $[e_1,e_2]=0$ we get the trivial bracket, and if it is non-zero we may, after a change of basis we may assume that $[e_1,e_2]=e_2$. The bracket is fully determined by this and it is easy to check that it satisfies condition 2 above and hence form a valid Lie algebra.
When we view these as abstract Lie algebras, no, not that I know of. It is algebraically one possible choice of bracket. If you want to interpret this as vector fields on a Lie group, then you might find a geometric interpretation somewhere.
Representations: If you are happy with the fact that there is a single 1-dimensional and two 2-dimensional Lie algebras up to isomorphism then the above is a fine classification. As I understand your question, you want to find a realization as a Lie subalgebra of $\mathfrak{gl}_2$ of $2\times 2$ matrices with the usual commutator bracket.
Warning: It is not true in general that a $n$-dimensional Lie algebra can be realized as a Lie subalgebra of $\mathfrak{gl}_n$.
For the abelian ones, any 1 and 2-dimensional sub vector space of commuting $2\times 2$ matrices will be isomorphic as a Lie algebra to the abelian ones we found (the bracket is zero on both of them so any linear isomorphism is a lie algebra isomorphism).
For the non-abelian one, let us denote it by $\mathfrak{g}$, we only need to find non-zero matrices $A_1,A_2$ such that $[A_1,A_2]=A_1A_2-A_2A_1=A_2$. Then the subalgebra $\mathfrak{h}$ of $\mathfrak{gl}_2$ spanned by these is isomorphic to $\mathfrak{g}$ via the map $\mathfrak{g}\to\mathfrak{h}$ where $e_i\mapsto A_i$. One choice here is
$$A_1=\begin{bmatrix}1 & 0 \\ 0 & 0 \end{bmatrix}, ~~ A_2=\begin{bmatrix}0 & 1 \\ 0 & 0 \end{bmatrix}.$$
You can find this by trial and error, or try to find a solution to the 4 equations you get from $[A_1,A_2]=A_2$.