= Solution
Start with the <approximate surjectivity with geometric correction> argument. For a fixed $y$, put $r_0=y$. Choose $x_j$ successively so that
$$
\|x_j\|\le R\|r_{j-1}\|,\qquad r_j=r_{j-1}-Tx_j,\qquad \|r_j\|\le k\|r_{j-1}\|.
$$
Induction gives $\|r_j\|\le k^j\|y\|$ and $\|x_j\|\le Rk^{j-1}\|y\|$. The series $\sum_jx_j$ is absolutely convergent in the <Banach space> $V$, because the sum of its norms is bounded by the convergent <geometric series> $R\|y\|/(1-k)$. Write its sum as $x$. The <bounded linear operator> $T$ commutes with this limit, while $T\sum_{j=1}^m x_j=y-r_m\to y$. Consequently
$$
\boxed{Tx=y,\qquad \|x\|\le\frac{R}{1-k}\|y\|.}
$$
The choices need not depend linearly on $y$; no bounded linear right inverse is being asserted.
For the bounded extension, use the restriction <bounded linear operator> $S:C_b(X;\mathbb R)\to C_b(Y;\mathbb R)$ on the <Banach spaces> of <bounded continuous functions>, each with its <supremum norm>. We will construct an approximation to any $g$ with $\|g\|_\infty=M>0$. The two sets
$$
A=\{y\in Y:g(y)\le -M/3\},\qquad B=\{y\in Y:g(y)\ge M/3\}
$$
are disjoint <closed sets> in $X$, since $Y$ is closed. If both are nonempty, the <distance to a set> construction defines a <continuous function>
$$
u(x)=\frac M3\frac{d(x,A)-d(x,B)}{d(x,A)+d(x,B)}.
$$
The denominator is positive at every point: zero distance to a closed set means membership in that set. Thus $\|u\|_\infty\le M/3$, with $u=-M/3$ on $A$ and $u=M/3$ on $B$. On $A$ and $B$, and also on the intervening range $|g|<M/3$, this implies
$$
\|Su-g\|_\infty\le 2M/3.
$$
If only $A$ is empty take $u=M/3$, if only $B$ is empty take $u=-M/3$, and if both are empty take $u=0$; the same bounds hold. For $M=0$ take $u=0$. This proves the approximation hypothesis with $R=1/3$ and $k=2/3$. Applying the argument above to $S$ yields an element $h$ of the <bounded continuous functions> with $h|_Y=g$ and $\|h\|_\infty\le\|g\|_\infty$. Restriction gives the reverse inequality, so \b[the extension preserves the exact supremum norm]:
$$
\boxed{\|h\|_\infty=\|g\|_\infty.}
$$
This proves the bounded metric-space case of the <Tietze extension theorem> rather than presupposing it.
For an arbitrary real <continuous function> $f$ on $Y$, let $g=\tanh f$. Its values lie strictly between $-1$ and $1$, although its supremum norm can be $1$. The bounded result supplies an extension $h$ with $|h|\le1$. Let $D=\{x:|h(x)|=1\}$, a <closed set> disjoint from $Y$. If $D$ is nonempty and $Y$ is nonempty, put
$$
\eta(x)=\frac{d(x,D)}{d(x,D)+d(x,Y)},\qquad q(x)=\eta(x)h(x).
$$
Both factors are <continuous functions>; $\eta=1$ on $Y$ and $\eta=0$ on $D$. Off $D$ we have $|h|<1$, and on $D$ we have $q=0$, so $|q(x)|<1$ everywhere. Therefore \b[a real continuous extension is]
$$
\boxed{F(x)=\operatorname{arctanh}q(x).}
$$
On $Y$ it equals $f$. If $D$ is empty use $q=h$; if $Y$ is empty simply take $F=0$. This handles unbounded $f$ without requiring a uniform margin between $|g|$ and $1$.
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