Step 1: Baire Category Argument

Let $B_X(r) = \{x \in X : \|x\| < r\}$ denote the open ball of radius $r$ in $X$. Since $T$ is surjective, we have $Y = T(X) = T\left(\bigcup_{n=1}^\infty B_X(n)\right) = \bigcup_{n=1}^\infty T(B_X(n))$

Since $Y$ is a complete metric space, the Baire Category Theorem implies that at least one $T(B_X(n_0))$ has non-empty interior. Therefore $T(B_X(1))$ also has non-empty interior (by linearity).

Let $U$ be an open set in $Y$ contained in $T(B_X(1))$. Then $0 \in U - U \subset T(B_X(1)) - T(B_X(1)) = T(B_X(2))$.

Thus there exists $\delta > 0$ such that $B_Y(\delta) \subset \overline{T(B_X(2))}$.

Step 2: Key Estimate

We claim: $B_Y(\delta) \subset T(B_X(4))$.

Let $y \in B_Y(\delta)$. Since $y \in \overline{T(B_X(2))}$, there exists $x_1 \in B_X(2)$ such that $\|y - Tx_1\| < \delta/2$.

Now $y - Tx_1 \in B_Y(\delta/2) \subset \overline{T(B_X(1))}$, so there exists $x_2 \in B_X(1)$ with $\|(y - Tx_1) - Tx_2\| < \delta/4$.

Continuing inductively, we find $x_n \in B_X(2^{1-n})$ such that $\left\|y - T\left(\sum_{k=1}^n x_k\right)\right\| < \delta 2^{-n}$

The series $\sum_{k=1}^\infty x_k$ converges in $X$ (by completeness) since $\sum_{k=1}^\infty \|x_k\| < 2 + 1 + 1/2 + \cdots = 4$

Let $x = \sum_{k=1}^\infty x_k$. Then $\|x\| < 4$ and by continuity of $T$: $Tx = T\left(\lim_{n \to \infty} \sum_{k=1}^n x_k\right) = \lim_{n \to \infty} T\left(\sum_{k=1}^n x_k\right) = y$

Therefore $y \in T(B_X(4))$.

Step 3: Conclusion

For any open set $U \subset X$, we show $T(U)$ is open. Let $y_0 = T(x_0) \in T(U)$ where $x_0 \in U$. Choose $r > 0$ such that $B_X(x_0, r) \subset U$.

From Step 2, $B_Y(\delta r/4) \subset T(B_X(r)) = T(B_X(x_0, r) - x_0)$. Therefore: $y_0 + B_Y(\delta r/4) = T(x_0) + T(B_X(r)) \subset T(U)$

Thus $B_Y(y_0, \delta r/4) \subset T(U)$, showing that $T(U)$ is open.