I define subspace to be a subset
That's incorrect. A subspace of a vector space $V$ is a subset $W\subseteq V$ which is also a vector space in its own right (with the addition and scalar multiplication "coming from $V$" in the obvious way). This means that $W$ must be
closed under addition (for each $a,b\in W$ we also have $a+b\in W$),
closed under scalar multiplication (for each scalar $s$ and each $a\in W$ we also have $s \cdot a\in W$), and
contain the zero vector,
where "$+$," "$\cdot$," and "zero vector" all take their meaning from $V$.
This more nuanced meaning of "sub[thing]" is ubiquitous in mathematics; compare "subgroup," "subring," ...
For example, consider $\mathbb{R}^2$ as a vector space over $\mathbb{R}$ in the usual way. The subset $$\{(0,r):r\in\mathbb{R}\}$$ is a subspace of $\mathbb{R}^2$, but $\{(1,r): r\in\mathbb{R}\}$ is not (in fact the latter fails all three of the bulletpoints above). The set $L(V,V)$ of your question is more difficult to picture, but the point is the same: the "zero vector" of $L(V,V)$ is the constant map $${\bf v}\mapsto{\bf 0}_V,$$ but this is usually not an isomorphism from $V$ to itself.
Exercise: when is the zero map an isomorphism?
Isomorphisms have to be bijective, so the zero map $V\rightarrow V$ is an isomorphism only if $V$ has exactly one element. And it's easy to check that in this case the zero map is indeed an isomorphism.