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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.