Show that space of all linear transformations L(V,V) from $V \to V$ is a vector space.
Vector spaces have 8 characteristics that describe them as vector spaces. Therefore, I think that to check if $L(V,V)$ is a vector space I need to check if it meets all the points.
Take as the basis of $V$ $t_1,t_2,...,t_n$
1) Given any two transformations $T:V \to V$ and $U: V\to V$ we must show that for any $v \in V$ $T(v) + U(v) \in L(V,V)$.
Since $T: V\to V$ $\exists v_1 \in V$ such that $T(v) = v$. The same for $U: V\to V$ $\exists v_2 \in V$ such that $U(v) = v_2$. Therefore, $T(v) + U(v) = v_1 + v_2$. Since $V$ is itself a vector space, then $v_1+v_2 \in V$, so $T(v) + U(v) \in V$. Also we have that $T(v) + U(v) = (T+U)v$ is linear. Therefore, $T(v) + U(v) \in L(V,V)$.
2) Given any transformation $T:V \to V$ and $a \in K$ we must show that for any $v \in V$ $kT \in L(V,V)$.
For any $v \in V$. Since $T: V\to V$ $\exists v_1 \in V$ such that $T(v) = v_1$. Since $T$ is a linear transformation by default then $kT(v) = kv_1$. Since $V$ is itself a vector space then $kv_1\in V$. Hence, $kT \in L(V,V)$.
3) Given any two transformations $T:V \to V$ and $U: V\to V$ we must show that for any $v \in V$ $T(v) + U(v) = v_1 + v_2$ and $U(v) + T(v) = v_2 + v_1$. Since $v_1 + v_2 = v_2 + v_1$ because of $V$ being linear space $T(v) + U(v) = U(v) + T(v)$
I guess, the other points will be done in a similar manner. Is this approach right? Are there any mistakes in the logic?
Before you start to prove each of the properties that define a vector space, it is essential to say why the sum and the scalar multiplication are well-defined there (which is what you tried to do). That is, if $\textsf T$ and $\textsf U$ are in $\mathcal{L}(\textsf V,\textsf V)$ then the functions $\textsf T + \textsf U$ and $a\textsf T$ are both in $\mathcal{L}(\textsf V,\textsf V)$, for any scalar $a$.
What do you have to do to see that? It's easy, you have to prove that these functions are linear. See, fix $x$ and $y$ vectors in $\textsf V$ and, for any scalar $c$, we have that $$\begin{align} (\textsf T + \textsf U)(cx+y) &=\textsf T(cx+y)+\textsf U(cx+y) \tag{by definition}\\ &=\big(c\textsf T(x)+\textsf T(y)\big)+\big(c\textsf U(x)+\textsf U(y)\big) \tag{$\textsf T$ and $\textsf U$ are linear}\\ &=c\big( \textsf T(x)+\textsf U(x) \big)+\big( \textsf T(y)+\textsf U(y) \big) \tag{...}\\ &=c(\textsf T + \textsf U)(x)+(\textsf T + \textsf U)(y) \tag{by definition, again} \end{align}$$ that is, $\textsf T + \textsf U$ is linear. Or better said, that $\textsf T + \textsf U \in \mathcal{L}(\textsf V,\textsf V)$. Now, do the same with $a\textsf T$.
Now that we prove that vector space operations are well-defined, we have every right to prove the $8$ properties, namely :
$(\textrm{i})$ : $\textsf T + (\textsf U+\textsf S)=(\textsf T + \textsf U) + \textsf S$ for any $\textsf{T},\textsf{U}$ and $\textsf S$ in $\mathcal{L}(\textsf V,\textsf V)$.
$(\textrm{ii})$ : $\textsf T + \textsf U=\textsf U + \textsf T$ for any $\textsf{T}$ and $\textsf{U}$ in $\mathcal{L}(\textsf V,\textsf V)$.
$(\textrm{iii})$ : There exists $\textit{0} \in \mathcal{L}(\textsf V,\textsf V)$ such that $\textsf T + \textit 0=\textsf T$ for all $\textsf T$ in $\mathcal{L}(\textsf V,\textsf V)$.
$(\textrm{iv})$ : For every $\textsf T \in \mathcal{L}(\textsf V,\textsf V)$ there exists $\textsf U \in \mathcal{L}(\textsf V,\textsf V)$ such that $\textsf T + \textsf U=\textit 0$.
$(\textrm{v})$ : $1\textsf T=\textsf T$ for each $\textsf T \in \mathcal{L}(\textsf V,\textsf V)$ (here, the $1$ denote the identity element of the field in question).
$(\textrm{vi})$ : $a(b\textsf T)=(ab)\textsf T$ for any scalars $a$ and $b$, and $\textsf T \in \mathcal{L}(\textsf V,\textsf V)$.
$(\textrm{vii})$ : $a(\textsf T + \textsf U)=a\textsf T+a\textsf U$ for any $\textsf{T}$ and $\textsf{U}$ in $\mathcal{L}(\textsf V,\textsf V)$ and for any scalar $a$.
$(\textrm{viii})$ : $(a+b)\textsf T=a\textsf T+b\textsf T$ for any scalars $a$ and $b$, and $\textsf T \in \mathcal{L}(\textsf V,\textsf V)$.
Those property are too easily to prove. For instance I'll help you with $(\textrm{iii})$, $(\textrm{iv})$ and $(\textrm{vii})$.
For $(\textrm{iii})$, define $\textit 0 : \textsf V\to\textsf V$ by $\textit 0(x)=0_\textsf{V}$, where $0_\textsf{V}$ denote the zero vector in $\textsf V$. It is clear that $\textit 0$ is a linear transformation, thus $\textit{0} \in \mathcal{L}(\textsf V,\textsf V)$. Now, for every $\textsf T \in \mathcal{L}(\textsf V,\textsf V)$ and $x\in \textsf V$ we have that $$(\textsf T+\textit 0)(x)=\textsf T(x)+\textit 0(x)=\textsf T(x)+0_\textsf{V}=\textsf T(x)$$ so, $\textsf T + \textit 0=\textsf T$ as we want to show.
A similar argument for $(\textrm{iv})$ will help us. Let $\textsf T \in \mathcal{L}(\textsf V,\textsf V)$. Define $\textsf U : \textsf V\to\textsf V$ by $\textsf U(x)=-\textsf T(x)$. Again, it is clear that $\textsf U \in \mathcal{L}(\textsf V,\textsf V)$ and observe that, for all $x\in \textsf V$, $$(\textsf T+\textsf U)(x)=\textsf T(x)+\textsf U(x)=\textsf T(x)-\textsf T(x)=0_\textsf{V}=\textit 0(x)$$ thus $\textsf T + \textsf U=\textit 0$.
And finally, for $(\textrm{vii})$, we have to do the same thing we are doing, just be careful, $a(\textsf T + \textsf U)$ and $a\textsf T+a\textsf U$ are the name of the functions, to prove their are in fact equal, you have to prove that $$\big( a(\textsf T + \textsf U) \big)(x)=\big( a\textsf T+a\textsf U \big)(x)$$ for all $x\in \textsf V$. To do so, fix an arbitrary $x\in \textsf V$, then $$\begin{align} \big( a(\textsf T + \textsf U) \big)(x) &= a (\textsf T+\textsf U)(x) \tag{vii$_1$}\\ &= a\big( \textsf T(x)+\textsf U(x) \big) \tag{vii$_2$}\\ &= a\textsf T(x)+a\textsf U(x) \tag{vii$_3$} \\ &= (a\textsf T)(x)+(a\textsf U)(x) \tag{vii$_4$}\\ &= \big( a\textsf T+a\textsf U \big)(x) \tag{vii$_5$} \end{align}$$ and then, we can conclude that $a(\textsf T + \textsf U)=a\textsf T+a\textsf U$, as we want to show. Can you explain what exactly happens in this previous steps?