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Put $C=R/(1-k)$. The hypothesis scales to every residual $r\in F$: if $r\ne 0$, apply it to $r/\Vert r\Vert$ and multiply the resulting vector by $\Vert r\Vert$ to obtain an approximate preimage of norm at most $R\Vert r\Vert$ with error at most $k\Vert r\Vert$. For $r=0$, choose zero.

Start with $r_0=y$. Inductively choose $u_j\in E$ so thatThen $\Vert r_j\Vert \le k^j\Vert y\Vert$ and ${\sum }_j\Vert u_j\Vert \le R\Vert y\Vert /(1-k)$. Completeness of the Banach space $E$ gives $x={\sum }_j{u}_j$. Boundedness of $T$ givesThusThis 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 $\Vert y_{n_{j+1}}-y_{n_j}\Vert \le 2^{-j}$. Lift $y_{n_1}$ to $v_0\in E$ and each difference to $v_j\in E$ with $T{v}_j=y_{n_{j+1}}-y_{n_j}$ and $\Vert v_j\Vert \le C{2}^{-j}$. The series $v_0+{\sum }_j{v}_j$ 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 $F$ is complete.

The open mapping theorem 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 $\Vert (x,y)\Vert =\Vert x\Vert +\Vert y\Vert$ 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; thereforeIt 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 $x_n\to 0$, $T{x}_n\to y$ and $z_n\to 0$, $T{z}_n\to w$, then $\alpha x_n+\beta z_n\to 0$ 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 $y_j$ withThen $w_j\to 0$ and $T{w}_j\to y$, 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 $x_n\to x$, $T{x}_n\to y$, then $x_n-x\to 0$ 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. ConsequentlyThe original PDF places $S$ in $F$; the TeX's first reference to a subset of $E$ is a transcription error.

First scale the stated unit-ball approximation. For every nonzero residual $r\in F$, apply the hypothesis to $r/\Vert r\Vert$ and multiply the resulting vector by $\Vert r\Vert$. This gives $v\in E$ withFor zero residual use $v=0$. Start with $r_0=y$, choose $v_j$ by this rule for $r_j$, and set $r_{j+1}=r_j-T{v}_j$. Inductively,The Banach space $E$ therefore contains the sum $x={\sum }_{j\ge 0}{v}_j$, withThe partial sums satisfy $T{\sum }_{j=0}^{n-1}{v}_j=y-r_n$. Boundedness of $T$ gives convergence of the left side to $Tx$, while $r_n\to 0$ gives $Tx=y$. Thus $T$ is surjective with the stated lifting bound. This geometric correction for approximate surjectivity used completeness of $E$, not any unproved completeness of $F$.

Put $C=R/(1-k)$. Given a Cauchy sequence $(y_n)$ in $F$, choose a subsequence $(y_{n_j})$ such that $\Vert y_{n_{j+1}}-y_{n_j}\Vert \le 2^{-j}$ for $j\ge 1$. Lift each difference to $u_j\in E$ with $T{u}_j=y_{n_{j+1}}-y_{n_j}$ and $\Vert u_j\Vert \le C{2}^{-j}$. Lift $y_{n_1}$ to $u_0$. The sum $u=u_0+{\sum }_{j\ge 1}{u}_j$ exists in $E$, and its image is the limit of the subsequence. A Cauchy sequence with a convergent subsequence converges to the same limit: use the triangle inequality between an arbitrary late term, a later subsequence term, and the limit. This proves $F$ is complete, the completeness forced by uniformly bounded lifting principle.

The open mapping theorem states that a bounded surjective linear operator between Banach spaces maps open sets to open sets. In particular a bounded linear bijection between Banach spaces has a bounded inverse. To deduce the closed graph theorem, let $S:E_1\to F_1$ be an everywhere-defined linear map between Banach spaces with closed graph. Its graph is a closed subspace of the Banach product $E_1\times F_1$, with norm $\Vert (x,y)\Vert =\Vert x\Vert +\Vert y\Vert$, so it is itself Banach. The projectionis a bounded linear bijection. The open mapping theorem makes its inverse bounded. Composing that inverse with the bounded second projection proves that $S$ is bounded. This is the required closed graph conclusion, rather than a continuity assumption on $S$.

Finally consider the identity map $I:(C(X),\Vert \cdot \Vert )\to (C(X),\Vert \cdot {\Vert }_{\infty })$. If $f_n\to f$ in the new norm and $f_n\to g$ uniformly, each assumed continuous point evaluation functional gives $f_n(x)\to f(x)$, while uniform convergence gives $f_n(x)\to g(x)$. Hence $f=g$. Since both spaces are metric, this sequential argument proves the graph is closed. Both norms are complete, so the closed graph theorem makes $I$ bounded. Its inverse is bounded by the open mapping theorem. Consequently there are constants $c,C^{\prime }>0$ such thatThis proves Banach norm rigidity from continuous point evaluations. Pointwise continuity alone would not give a common bound on all evaluations; completeness supplies that through the closed graph argument.