The Set of Complex Numbers as a Vector Space
Explanation
The set of complex numbers $\mathbb{C}$ is itself a field, and it becomes a $\mathbb{C}$-vector space with itself as the scalar field. That is, every $z \in \mathbb{C}$ can be uniquely expressed as $z = z \cdot 1$, so it is a $1$-dimensional vector space with basis $\left\{ 1 \right\}$.
$$ \dim_{\mathbb{C}}(\mathbb{C}) = 1,\quad \mathbb{C} = \span_{\mathbb{C}} \left\{ 1 \right\} $$
However, if we take the real numbers $\mathbb{R}$ as the scalar field, the result changes. Every $z$ can be uniquely expressed as $z = x + iy$ for two real numbers $x, y \in \mathbb{R}$. Hence in this case the basis that spans $\mathbb{C}$ is $\left\{ 1, i \right\}$.
$$ \dim_{\mathbb{R}}(\mathbb{C}) = 2,\quad \mathbb{C} = \span_{\mathbb{R}} \left\{ 1, i \right\} $$
In this way, even the same set can have a different vector space structure depending on which scalar field it is viewed over. When the scalar field is changed from $\mathbb{C}$ to $\mathbb{R}$, the size of the basis grows from $1$ to $2$, but this is not because new elements appeared that did not exist before. The set $\mathbb{C}$ stays the same; only the judgment of linear independence for the same two elements $1$ and $i$ is reversed depending on the scalar field. Since $i = i \cdot 1$, $\left\{ 1, i \right\}$ is linearly dependent over $\mathbb{C}$, but for $a, b \in \mathbb{R}$, if $a \cdot 1 + b \cdot i = 0$ then $a = b = 0$, so it is linearly independent over $\mathbb{R}$. In other words, whether something is linearly independent itself depends on the scalar field.
Increasing the dimension and looking at $\mathbb{C}^{2}$, we get the following difference.
$$ \dim_{\mathbb{C}}(\mathbb{C}^{2}) = 2,\quad \mathbb{C}^{2} = \span_{\mathbb{C}} \left\{ (1,0), (0,1) \right\} $$
$$ \dim_{\mathbb{R}}(\mathbb{C}^{2}) = 4,\quad \mathbb{C}^{2} = \span_{\mathbb{R}} \left\{ (1,0), (i,0), (0,1), (0,i) \right\} $$
In general, since an $n$-dimensional real vector space is isomorphic to $\mathbb{R}^{n}$, we obtain the following result for $\mathbb{C}^{n}$ as an $\mathbb{R}$-vector space.
$$ \mathbb{C}^{n} \cong \mathbb{R}^{2n} $$
Subspaces
Since $\mathbb{C}$ as a $\mathbb{C}$-vector space is $1$-dimensional, its only subspaces are $\left\{ 0 \right\}$ and $\mathbb{C}$ itself. The set of real numbers might seem to be a subspace, but it is not. Since $i \cdot 1 = i \notin \mathbb{R}$, $\mathbb{R}$ is not closed under multiplication by complex numbers, so it is not a $\mathbb{C}$-subspace.
On the other hand, viewed as an $\mathbb{R}$-vector space, for each $z_{0} \ne 0$ as below, every line through the origin is a $1$-dimensional $\mathbb{R}$-subspace.
$$ \mathbb{R}z_{0} = \left\{ t z_{0} : t \in \mathbb{R} \right\} \subset \mathbb{C}, \quad \forall z_{0} \in \mathbb{C} $$
Thus the $\mathbb{R}$-subspaces of $\mathbb{C}$ are infinitely many: $\left\{ 0 \right\}$, the lines through the origin (including $\mathbb{R}$), and $\mathbb{C}$ itself. Therefore, speaking of $\mathbb{R}$ as a subspace of $\mathbb{C}$ means assuming that the scalar field is taken to be the set of real numbers.