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
$$\varphi(\mathfrak{h})=\left\{\left(\begin{array}{ll}
a&b\\
0&0
\end{array}\right)\in\mathfrak{gl}(2,\mathbb{R}):a,b\in\mathbb{R}\right\}\subset\mathfrak{gl}(2,\mathbb{R}).$$
Best Answer
Yes, this is true.
First approach
Let more generally $(\mathfrak{g}, [-,-]_1)$ be a Lie algebra over some field $K$. If $\mathfrak{h}$ is a $K$-vector space of the same dimension as $\mathfrak{g}$ then there exists an isomorphism of vector spaces $$ \varphi \colon \mathfrak{h} \to \mathfrak{g} \,. $$ We can use this isomorphism of vector spaces to pull back the Lie bracket on $\mathfrak{g}$ to a Lie bracket on $\mathfrak{h}$. More precisely, we set $$ [x, y]_2 := \varphi^{-1}( [ \varphi(x), \varphi(y) ]_1 ) $$ for all $x, y \in \mathfrak{h}$. This defines a Lie bracket on $\mathfrak{h}$, and the vector space isomorphism $\varphi$ becomes an isomorphism of Lie algebras $$ \varphi \colon ( \mathfrak{h}, [-,-]_2 ) \to ( \mathfrak{g}, [-,-]_1 ) \,. $$ (This Lie bracket $[-,-]_2$ is the unique Lie bracket on $\mathfrak{h}$ which makes $\varphi$ into an isomorphism of Lie algebras.)
Suppose now that $K = \mathbb{R}$ and that the Lie algebra $\mathfrak{g}$ is of finite dimension $n$. If $x_1, \dotsc, x_n$ is a basis of $\mathfrak{g}$ then we get an isomorphism of vector spaces $$ \varphi \colon \mathbb{R}^n \to \mathfrak{g} \,, \quad e_i \mapsto x_i \qquad \text{for all $i = 1, \dotsc, n$,} $$ where $e_1, \dotsc, e_n$ denotes the standard basis of $\mathfrak{g}$. As explained above we can pull back the Lie bracket $[-,-]_1$ on $\mathfrak{g}$ to a Lie bracket $[-,-]_2$ on $\mathbb{R}^n$ such that $\varphi$ becomes an isomorphism of Lie algebras.
Second approach
There is also a less abstract but equivalent way of explaining the above procedure. Suppose again that $(\mathfrak{g}, [-,-]_1)$ is a Lie algebra over a field $K$ and let $x_1, \dotsc, x_n$ be a basis of $\mathfrak{g}$. We can then consider the structure constants of $[-,-]_1$ with respect to the basis $x_1, \dotsc, x_n$. These are the unique coefficients $c_{ij}^k \in K$ with $$ [x_i, x_j] = \sum_{k=1}^n c_{ij}^k x_k $$ for all $i, j = 1, \dotsc, n$. That $[-,-]_1$ is a Lie bracket on $\mathfrak{g}$ is equivalent to the conditions \begin{equation} c_{ii}^\ell = 0 \,, \quad c_{ij}^\ell + c_{ji}^\ell = 0 \,, \quad \sum_{k=1}^n ( c_{ij}^k c_{k \ell}^m + c_{j \ell}^k c_{ki}^m + c_{\ell i}^k c_{kj}^m ) = 0 \tag{1} \end{equation} for all $i, j, \ell, m = 1, \dotsc, n$. If we are now given another $K$-vector space $\mathfrak{h}$ with basis $y_1, \dotsc, y_n$ then we can define a unique bilinear map $$ [-,-]_2 \colon \mathfrak{h} \times \mathfrak{h} \to \mathfrak{h} $$ such that $$ [y_i, y_j] := \sum_{k=1}^n c_{ij}^k y_k $$ for all $i,j = 1, \dotsc, n$. The structure constants of $[-,-]_2$ with respect to the basis $y_1, \dotsc, y_n$ of $\mathfrak{h}$ are (by construction) the same as the structure constants of $[-,-]_1$ with respect to the basis $x_1, \dotsc, x_n$ of $\mathfrak{g}$. It follows that $[-,-]_2$ is a Lie bracket on $\mathfrak{h}$ since the conditions $(1)$ are satisfied. Moreover, there exists a unique linear map $\varphi$ from $\mathfrak{h}$ to $\mathfrak{g}$ with $$ \varphi(y_i) = x_i $$ for all $i = 1, \dotsc, n$, and this linear map is an isomorphism of Lie algebras from $(\mathfrak{h}, [-,-]_2)$ to $(\mathfrak{g}, [-,-]_1)$.
In the given situation we have $K = \mathbb{R}$ and can choose $\mathfrak{h} = \mathbb{R}^n$ and for $y_1, \dotsc, y_n$ the standard basis of $\mathbb{R}^n$.