Let’s simplify the situation for a moment. Forget that $V$ and $U$ are vector spaces and that $T$ is a linear transformation; just think of $V$ and $U$ as sets and of $T$ as a map from $V$ onto $U$. For each $u\in U$ let $V_u=\{v\in V:T(v)=u\}$, the set of points in $V$ that are mapped to $u$ by $T$. Let $\mathscr{P}=\{V_u:u\in U\}$. If $u_0,u_1\in U$ and $u_0\ne u_1$, then $V_{u_0}\cap V_{u_1}=\varnothing$: $T$ cannot send any $v\in V$ both to $u_0$ and to $u_1$. Thus, the map $\varphi:U\to\mathscr{P}:u\mapsto V_u$ is a bijection, and so, of course, is its inverse sending $V_u\in\mathscr{P}$ to $u$.
Now put the linear algebra back into the picture. First, $\ker T=\{v\in V:T(v)=0_U\}$, so in the notation of my first paragraph, $\ker T=V_{0_U}$: it’s one of the members of $\mathscr{P}$. Fix $v_0\in V$ and let $u_0=T(v_0)$; what vectors in $V$ belong to $V_{u_0}$? Suppose that $v\in V_{u_0}$; then $T(v)=u_0=T(v_0)$. Since $T$ is linear, $T(v-v_0)=T(v)-T(v_0)=0_U$, so $v-v_0\in\ker T$, and $v\in v_0+\ker T$, where $v_0+\ker T=\{v_0+v:v\in\ker T\}$. Conversely, you can easily check that if $v\in v_0+\ker T$, then $T(v)=u_0$, and therefore $v\in V_{u_0}$. Thus, $V_{u_0}=v_0+\ker T$. In other words, the members of $\mathscr{P}$ are precisely the sets of the form $v_0+\ker T$ for $v_0\in V$.
By definition the members of $V/\ker T$ are the sets $v_0+\ker T$ for $v_0\in V$, and we’ve just seen that these are the members of $\mathscr{P}$, so in fact $V/\ker T=\mathscr{P}$. Thus, we can just as well think of the map $\varphi$ defined above as a bijection from $U$ onto $V/\ker T$. Its inverse, which I’ll call $h$, is a bijection from $V/\ker T$ onto $U$. What does $h$ look like? Let $v_0+\ker T\in V/\ker T$, and let $u_0=T(v_0)$. We’ve just seen that $v_0+\ker T=V_{u_0}$, and we know from the first paragraph that $h(V_{u_0})=u_0$. In other words, $h(v_0+\ker T)=u_0=T(v_0)$.
We’ve now shown that the map $h:V/\ker T\to U:v+\ker T\mapsto T(v)$ is a bijection; in terms of the sets involved, it’s just the inverse of the bijection $\varphi$ of the first paragraph. To complete the proof that $V/\ker T$ and $U$ are isomorphic, we just check that $h$ is linear, which is a straightforward computation.
You cannot prove it, since it is not true. If it were true, then any injective operator between Banach spaces would be continuous.
Take any two infinite-dimensional Banach spaces with the same dimension (as vector spaces). Then there is a linear isomorphism $L$ between them, whose kernel is closed, since it is $\{0\}$. But, since they are not isomorphic, either $L$ is not continuous or $L^{-1}$ is not continuous. And $\ker L^{-1}=\{0\}$ too.
Best Answer
This is not true in general:
Let $X$ be any infinite-dimensional banach space with (Hamel) basis $B$ and let $\{b_n\}_{n\in\ \mathbb N}$ be a countable subset of $B$. Then define $T:X\to X$ on the basis $B$ by $T(b_n)=nb_n$ and $T(b)=b$ for $b\in B\setminus \{b_n\}_{n\in\ \mathbb N}$ and extend linearly. Then $T$ is an unbounded operator, with $\ker T=\{0\}$ closed.