"Suppose $V$ and $W$ are finite-dimensional with $\dim V \ge \dim W \ge 2$. Show that $U=\{T \in \mathcal{L}(V, W): T $ is not surjective $\}$ is not a subspace of $\mathcal{L}(V, W)$. "
( $\mathcal{L}(V, W)$ is the set of all linear maps from $V$ to $W$ )
My attempt: If I assume that $U$ is a subspace of $\mathcal{L}(V, W)$ then
Let $T,S \in U$, $v\in V$ and let $ T(v) = w_1, S(v) = w_2 $
$$(T+S)(v)=T(v)+S(v) = w_1 + w_2 \longrightarrow T + S \in U$$ But this cannot be true because $w_1 \in \operatorname{range}(T)$ and $w_2 \in \operatorname{range}(S)$: $$ \operatorname{range}(T) \neq \operatorname{range}(S)$$
$T+S\notin U$, $U$ is not a subspace.
Is this proof correct?
My second attempt:If I assume that $U$ is a subspace of $\mathcal{L}(V, W)$ then
Let $T,S \in U$, $v\in V$ and let $ T(v) = w_1, S(v) = w_2 $
$$(S+T)(v)=S(v)+T(v)=w_1+w_2$$ $w_1 \in \operatorname{range}(T)$ and $w_2 \in \operatorname{range}(S)$
We know that the range of any linear map $T \in \mathcal{L}(V, W)$ is a subspace of $W$. So in this case, $\operatorname{range}(T)$ and $\operatorname{range}(S)$ are subspaces of W. This means that $$\operatorname{span}(\operatorname{range}(T))=\operatorname{span}(u_1,u_2,\dots,u_m)$$ $$\operatorname{span}(\operatorname{range}(S))=\operatorname{span}(u_{m+1},u_{m+2},\dots,u_n)$$ $u_j \in W, j=1,2,\dots,n$
$w_1$ and $w_2$ can be expressed as linear combinations $$ w_1 = k_1u_1 + k_2u_2 + \dots + k_mu_m$$ $$ w_2 = k_{m+1}u_{m+1} + k_{m+2}u_{m+2} + \dots + k_nu_n$$
Because $u_1,u_2,\dots, u_m $ is a basis of $ \operatorname{range}(T)$, there exists a unique linear map such that $$T(v_j) = u_j$$ $v_j \in V, u_j \in W, j=1,2\dots,m$
Likewise, because $u_{m+1},u_{m+2},\dots,u_{n}$ is a basis of $\operatorname{range}(S)$, there exists a unique linear map such that $$S(v_j) = u_j $$ $v_j \in V, u_j \in W, j=m+1,m+2,\dots, n$
This means that there are no $v_j \in V$ such that $$ T(v_j) = S(v_j) $$ If there would exist $T,S \in U$ where $ T(v_j) = S(v_j)$ then $$k_1u_1 + k_2u_2 + \dots + k_mu_m=w_2 = k_{m+1}u_{m+1} + k_{m+2}u_{m+2} + \dots + k_nu_n$$
$$\operatorname{range}(T) = \operatorname{range}(S) $$ But this is not the case. Therefore $$ \operatorname{range}(T) \ne \operatorname{range}(S) $$
Because $\operatorname{range}(T) \ne \operatorname{range}(S)$, there exists $w_j \in \operatorname{range}(T), w_j \notin \operatorname{range}(S) $. Let $j=1$. Now we have
$$ w_1 \in \operatorname{range}(T), w_1 \notin \operatorname{range}(S) $$ $w_2$ can either exist both in $\operatorname{range}(S)$ and $\operatorname{range}(T)$ or only in $\operatorname{range}(S)$. We test both alternatives and see the result.
Let $w_2 \in \operatorname{range}(S), w_2 \in \operatorname{range}(T)$
$$w_1+w_2 \in \operatorname{range}(T)$$ $$w_1+w_2 \notin \operatorname{range}(S)$$ This means that the range of T is closed under addition but the range of S is not closed under addition. $\operatorname{range}(S)$ can therefore not be a subspace of W. That means that $S \notin \mathcal{L}(V, W)$.
If $U=\{T \in \mathcal{L}(V, W): T $ is surjective $\}$, then $\operatorname{range}(T) = \operatorname{range}(S) = W$. But in this case $\operatorname{range}(T) \ne \operatorname{range}(S) \ne W$
Attempt 3: We know that $$\operatorname{dim}(V) \ge \operatorname{dim}(W) \ge 2$$ We know that $U=\{T \in \mathcal{L}(V, W): T $ is not surjective $\}$.
Let $T \in U$. $T$ is not surjective, which implies that $$ \operatorname{range}(T) \ne W$$ But because $ T \in \mathcal{L}(V, W) $, $\operatorname{range}(T)$ is a subspace of $W$. Let $w_1, w_2, w_3, ... , w_n$ span $\operatorname{range}(T)$. $$\operatorname{span}(\operatorname{range}(T)) = \operatorname{span}(w_1, w_2, \dots, w_n)$$ Now because $\operatorname{range}(T)$ is a subspace of $W$. $w_1, w_2,\dots, w_n$ can be expanded to span $W$. $$ \operatorname{span}(W) = \operatorname{span}(w_1, w_2, \dots , w_n, w_{n+1}, w_{n+2}, \dots , w_{m}) $$ This implies that $\operatorname{dim}(\operatorname{range}(T)) < \operatorname{dim}(W)$, which means that $$ \operatorname{dim}(V) > \operatorname{dim}(\operatorname{range}(T)) \ge 2 $$ We can check if $U$ is injective by checking if $ \operatorname{null}(T) = \{ 0 \} $. In order for that to be true then $T(u) = T(v), u = v, u,v \in V$.
Let $$ \operatorname{span}(w_1, w_2, \dots , w_m) = \operatorname{span}(Tv_1, Tv_2, \dots , Tv_m), Tv_j = w_j, j=1,2\dots,m, v_j \in V, w_j \in W $$ This means that $$ T(v) = k_1w_1 + k_2w_2 + \dots + k_nw_n + k_{n+1}w_{n+1} + \dots + k_mw_m = k_1Tv_1 + k_2Tv_2 + \dots + k_nTv_n + k_{n+1}Tv_{n+1} + \dots + k_mTv_m $$
But we know that $T(v) \in \operatorname{range}(T) $ and because of that $$ T(v) = k_1w_1 + k_2w_2 + \dots + k_nw_n = k_1 = k_1Tv_1 + k_2Tv_2 + \dots + k_nTv_n$$
This means that $$k_1Tv_1 + k_2Tv_2 + \dots + k_nTv_n = k_1Tv_1 + k_2Tv_2 + \dots + k_nTv_n + k_{n+1}Tv_{n+1} + \dots + k_mTv_m $$ $$ k_{n+1}Tv_{n+1} + \dots + k_mTv_m = 0 $$ $$T(k_{n+1}v_{n+1}) + \dots + T(k_mv_m) = 0 $$ $$T(k_{n+1}v_{n+1} + \dots + k_mv_m ) = 0$$ Which implies that $$ k_{n+1}v_{n+1} + \dots + k_mv_m \in \operatorname{null}(T)$$ This means that $\operatorname{null}(T) \ne \{0\} $ which means that T is not injective. So for all $T \in \mathcal{L}(V, W)$, $T$ is not injective.
Now, let $T,S \in \mathcal{L}(V, W) $ and $u,v \in V$ such that $T(v) = T(u), u \ne v $ $$ (T+S)(v) = T(v) + S(v) = T(u) + S(v) \ne (T+S)(v) $$
Pick a basis $v_1,...,v_m,...,v_n$ of $V$ and a basis $w_1,...,w_m$ of $W$. Consider the linear maps $Tv_i=w_i$, for $i=1,...,m-1$ and $Tv_i=0$, for $i\geq m$, while $Sv_m=w_m$ and $Sv_i=0$ for $i\neq m$. When you add them you get a surjective linear map.