= Solution
Put $C=R/(1-k)$. The hypothesis scales to every residual $r\in F$: if $r\ne0$, apply it to $r/\|r\|$ and multiply the resulting vector by $\|r\|$ to obtain an approximate preimage of <norm> at most $R\|r\|$ with error at most $k\|r\|$. For $r=0$, choose zero.
Start with $r_0=y$. Inductively choose $u_j\in E$ so that
$$
\|u_j\|\leq R\|r_{j-1}\|,\qquad r_j=r_{j-1}-Tu_j,\qquad \|r_j\|\leq k\|r_{j-1}\|.
$$
Then $\|r_j\|\leq k^j\|y\|$ and $\sum_j\|u_j\|\leq R\|y\|/(1-k)$. <Completeness> of the <Banach space> $E$ gives $x=\sum_j u_j$. Boundedness of $T$ gives
$$
\|Tx-y\|\leq\|T\|\left\|x-\sum_{j=1}^n u_j\right\|+\|r_n\|\longrightarrow0.
$$
Thus
$$
\boxed{Tx=y,\qquad \|x\|\leq\frac{R}{1-k}\|y\|}.
$$
This <geometric correction for approximate surjectivity> uses no <completeness> assumption on $F$: every infinite sum was formed in $E$.
To establish the <completeness forced by uniformly bounded lifting>, let $(y_n)$ be a <Cauchy sequence> in $F$. Choose a subsequence $(y_{n_j})$ with $\|y_{n_{j+1}}-y_{n_j}\|\leq2^{-j}$. Lift $y_{n_1}$ to $v_0\in E$ and each difference to $v_j\in E$ with $Tv_j=y_{n_{j+1}}-y_{n_j}$ and $\|v_j\|\leq C2^{-j}$. The series $v_0+\sum_jv_j$ converges in $E$, and its image under $T$ is the limit of the subsequence. The original <Cauchy sequence> has that same limit, by the triangle inequality. Hence \b[$F$ is complete].
The <open mapping theorem (functional analysis)> states that a bounded surjective <linear map> between <Banach spaces> is open. In particular, a bounded linear bijection has a bounded inverse. This is the theorem the question permits us to state without proof.
For the <closed graph theorem>, suppose $T:E\to F$ is everywhere-defined and linear with closed graph $G(T)$. The product <norm> $\|(x,y)\|=\|x\|+\|y\|$ makes $E\times F$ a <Banach space>, so its closed subspace $G(T)$ is complete. Projection $\pi:G(T)\to E$, $(x,Tx)\mapsto x$, is a bounded linear bijection. Its inverse is bounded by the <open mapping theorem (functional analysis)>; therefore
$$
\|x\|+\|Tx\|=\|\pi^{-1}x\|\leq C_1\|x\|.
$$
It follows that $T$ is bounded, proving the <closed graph theorem> rather than assuming it.
For the final clause, $S$ is the <separating space of a linear map>. It contains zero, and if $x_n\to0$, $Tx_n\to y$ and $z_n\to0$, $Tz_n\to w$, then $\alpha x_n+\beta z_n\to0$ and its image tends to $\alpha y+\beta w$. Thus $S$ is a <vector subspace>. If $y_j\in S$ and $y_j\to y$, choose one vector $w_j$ from a witnessing sequence for $y_j$ with
$$
\|w_j\|<1/j,\qquad \|Tw_j-y_j\|<1/j.
$$
Then $w_j\to0$ and $Tw_j\to y$, so $y\in S$: it is closed. If $T$ is continuous, every defining image limit is zero, hence $S=\{0\}$. Conversely, if $S=\{0\}$ and $x_n\to x$, $Tx_n\to y$, then $x_n-x\to0$ and $T(x_n-x)\to y-Tx\in S$, so $y=Tx$. The graph is sequentially closed and therefore closed in the normed, metrizable product. The proved <closed graph theorem> now gives continuity. Consequently
$$
\boxed{S\text{ is a closed linear subspace of }F,\qquad T\text{ is continuous}\iff S=\{0\}}.
$$
The original PDF places $S$ in $F$; the TeX's first reference to a subset of $E$ is a transcription error.
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