[proofplan]
We verify the algebra structure directly from the definitions. First we show that pointwise addition and scalar multiplication keep endomorphisms inside $\operatorname{End}_F(V)$ and give an $F$-[vector space](/page/Vector%20Space) structure. Then we prove that composition is a closed, associative multiplication with unit $\operatorname{id}_V$. Finally we check distributivity and scalar compatibility, which are exactly the remaining $F$-algebra axioms.
[/proofplan]
[step:Show that pointwise operations make $\operatorname{End}_F(V)$ an $F$-vector space]
Let $S,T \in \operatorname{End}_F(V)$ and $a \in F$. Define the map
\begin{align*}
A: V \to V,\quad v \mapsto S(v)+T(v).
\end{align*}
For $\lambda \in F$ and $v,w \in V$, using the $F$-linearity of $S$ and $T$ and the vector space operations in $V$, we have
\begin{align*}
A(\lambda v+w)=S(\lambda v+w)+T(\lambda v+w)=\lambda S(v)+S(w)+\lambda T(v)+T(w)=\lambda A(v)+A(w).
\end{align*}
Thus $A$ is $F$-linear, so $S+T=A \in \operatorname{End}_F(V)$.
Define also
\begin{align*}
B: V \to V,\quad v \mapsto aT(v).
\end{align*}
For $\lambda \in F$ and $v,w \in V$, using the $F$-linearity of $T$ and commutativity of multiplication in the field $F$,
\begin{align*}
B(\lambda v+w)=aT(\lambda v+w)=a(\lambda T(v)+T(w))=\lambda aT(v)+aT(w)=\lambda B(v)+B(w).
\end{align*}
Thus $B$ is $F$-linear, so $aT=B \in \operatorname{End}_F(V)$.
The zero map
\begin{align*}
0_{\operatorname{End}}: V \to V,\quad v \mapsto 0_V
\end{align*}
belongs to $\operatorname{End}_F(V)$, and for each $T \in \operatorname{End}_F(V)$ the map $-T: V \to V$ given by $v \mapsto -T(v)$ belongs to $\operatorname{End}_F(V)$. Since equality of maps is pointwise equality, every vector space axiom for $\operatorname{End}_F(V)$ follows by evaluating at an arbitrary $v \in V$ and using the corresponding vector space axiom in $V$. Therefore $\operatorname{End}_F(V)$ is an $F$-vector space.
[guided]
We first need to prove that the formulas in the statement actually define operations on $\operatorname{End}_F(V)$, not merely on all functions $V \to V$. This means that whenever $S$ and $T$ are $F$-linear maps from $V$ to $V$, their pointwise sum must again be an $F$-[linear map](/page/Linear%20Map) from $V$ to $V$.
Let $S,T \in \operatorname{End}_F(V)$ and define
\begin{align*}
A: V \to V,\quad v \mapsto S(v)+T(v).
\end{align*}
The codomain is $V$ because $S(v),T(v) \in V$ and $V$ is closed under vector addition. To check $F$-linearity, let $\lambda \in F$ and $v,w \in V$. Since $S$ and $T$ are $F$-linear,
\begin{align*}
S(\lambda v+w)=\lambda S(v)+S(w).
\end{align*}
\begin{align*}
T(\lambda v+w)=\lambda T(v)+T(w).
\end{align*}
Substituting these two identities into the definition of $A$ gives
\begin{align*}
A(\lambda v+w)=S(\lambda v+w)+T(\lambda v+w)=\lambda S(v)+S(w)+\lambda T(v)+T(w).
\end{align*}
Using associativity and commutativity of addition in the vector space $V$, we regroup the right-hand side as
\begin{align*}
\lambda S(v)+S(w)+\lambda T(v)+T(w)=\lambda(S(v)+T(v))+(S(w)+T(w))=\lambda A(v)+A(w).
\end{align*}
Hence $A$ is $F$-linear, so $S+T=A$ lies in $\operatorname{End}_F(V)$.
Now let $a \in F$ and define
\begin{align*}
B: V \to V,\quad v \mapsto aT(v).
\end{align*}
Again the codomain is $V$ because scalar multiplication by $a$ sends vectors of $V$ to vectors of $V$. For $\lambda \in F$ and $v,w \in V$, the linearity of $T$ gives
\begin{align*}
B(\lambda v+w)=aT(\lambda v+w)=a(\lambda T(v)+T(w)).
\end{align*}
Distributing scalar multiplication over vector addition in $V$ gives
\begin{align*}
a(\lambda T(v)+T(w))=a\lambda T(v)+aT(w).
\end{align*}
Since $F$ is a field, its multiplication is commutative, so $a\lambda=\lambda a$. Therefore
\begin{align*}
a\lambda T(v)+aT(w)=\lambda aT(v)+aT(w)=\lambda B(v)+B(w).
\end{align*}
Thus $B$ is $F$-linear, so $aT=B$ lies in $\operatorname{End}_F(V)$.
It remains in this step to justify the vector space axioms. The zero vector in $\operatorname{End}_F(V)$ is the zero map
\begin{align*}
0_{\operatorname{End}}: V \to V,\quad v \mapsto 0_V.
\end{align*}
This map is $F$-linear because $0_{\operatorname{End}}(\lambda v+w)=0_V=\lambda 0_V+0_V$. The additive inverse of $T$ is the map
\begin{align*}
-T: V \to V,\quad v \mapsto -T(v),
\end{align*}
which is $F$-linear by the scalar-multiple argument with scalar $-1_F$.
For the remaining axioms, such as associativity and commutativity of addition, distributivity of scalar multiplication over addition, and the scalar identity law, equality of maps reduces the verification to pointwise equality. For example, for $R,S,T \in \operatorname{End}_F(V)$ and $v \in V$,
\begin{align*}
((R+S)+T)(v)=(R(v)+S(v))+T(v)=R(v)+(S(v)+T(v))=(R+(S+T))(v).
\end{align*}
The middle equality is associativity of addition in $V$. The other vector space axioms are verified in the same pointwise way from the corresponding axioms in $V$. Therefore $\operatorname{End}_F(V)$ is an $F$-vector space.
[/guided]
[/step]
[step:Prove that composition is a closed associative multiplication with identity]
Let $S,T \in \operatorname{End}_F(V)$. Define
\begin{align*}
ST: V \to V,\quad v \mapsto S(T(v)).
\end{align*}
For $\lambda \in F$ and $v,w \in V$, using first the linearity of $T$ and then the linearity of $S$,
\begin{align*}
(ST)(\lambda v+w)=S(T(\lambda v+w))=S(\lambda T(v)+T(w))=\lambda S(T(v))+S(T(w))=\lambda(ST)(v)+(ST)(w).
\end{align*}
Thus $ST \in \operatorname{End}_F(V)$.
Let $R,S,T \in \operatorname{End}_F(V)$. For every $v \in V$,
\begin{align*}
((RS)T)(v)=(RS)(T(v))=R(S(T(v)))=R((ST)(v))=(R(ST))(v).
\end{align*}
Hence $(RS)T=R(ST)$, so multiplication is associative.
The identity map
\begin{align*}
\operatorname{id}_V: V \to V,\quad v \mapsto v
\end{align*}
is $F$-linear. For every $T \in \operatorname{End}_F(V)$ and $v \in V$,
\begin{align*}
(\operatorname{id}_V T)(v)=\operatorname{id}_V(T(v))=T(v).
\end{align*}
\begin{align*}
(T\operatorname{id}_V)(v)=T(\operatorname{id}_V(v))=T(v).
\end{align*}
Therefore $\operatorname{id}_V T=T=T\operatorname{id}_V$, so $\operatorname{id}_V$ is the multiplicative identity.
[/step]
[step:Verify distributivity and scalar compatibility]
Let $R,S,T \in \operatorname{End}_F(V)$. For every $v \in V$,
\begin{align*}
(R(S+T))(v)=R((S+T)(v))=R(S(v)+T(v))=R(S(v))+R(T(v))=(RS+RT)(v).
\end{align*}
Thus $R(S+T)=RS+RT$. Similarly, for every $v \in V$,
\begin{align*}
((R+S)T)(v)=(R+S)(T(v))=R(T(v))+S(T(v))=(RT+ST)(v).
\end{align*}
Thus $(R+S)T=RT+ST$.
Let $a \in F$ and $S,T \in \operatorname{End}_F(V)$. For every $v \in V$,
\begin{align*}
((aS)T)(v)=(aS)(T(v))=aS(T(v))=(a(ST))(v).
\end{align*}
Also, using the $F$-linearity of $S$,
\begin{align*}
(S(aT))(v)=S((aT)(v))=S(aT(v))=aS(T(v))=(a(ST))(v).
\end{align*}
Therefore
\begin{align*}
(aS)T=a(ST)=S(aT).
\end{align*}
[/step]
[step:Conclude the unital associative $F$-algebra structure]
The first step proves that $\operatorname{End}_F(V)$ is an $F$-vector space under pointwise addition and scalar multiplication. The second step proves that composition gives a closed associative multiplication with identity element $\operatorname{id}_V$. The third step proves that this multiplication is distributive over addition and compatible with scalar multiplication over $F$. These are precisely the axioms of a unital associative $F$-algebra, so $\operatorname{End}_F(V)$ is a unital associative $F$-algebra under the stated operations.
[/step]
