You can see the question in the title and my solution is written below:
Question: Let $T: X \rightarrow Y$ and $J: Y \rightarrow Z$ be linear, moreover suppose that $J$ is injective and bounded such that $J T: X \rightarrow Z$ is bounded as well. Show that $T$ is bounded.
My solution: To show that the linear operator $T: X \rightarrow Y$ is bounded, given that $J: Y \rightarrow Z$ is injective and bounded and the composition $J T: X \rightarrow Z$ is bounded, we'll use the Closed Graph Theorem. According to this theorem, if an operator is closed, then it is bounded. Therefore, we need to prove that $T$ is closed.
Let's consider a sequence $\left(x_n\right)$ in $X$ such that $x_n$ converges to some $x$ in $X$ and $T\left(x_n\right)$ converges to some $y$ in $Y$. We need to show that $T(x)=y$ to establish that $T$ is closed. Since $J T$ is bounded, it is continuous, and therefore $J T\left(x_n\right)$ converges to $J T(x)$. Also, as $J$ is continuous and bounded, $J\left(T\left(x_n\right)\right)$ converges to $J(y)$.
Now, since $J T\left(x_n\right)=J\left(T\left(x_n\right)\right)$ for each $n$, the sequences $J T\left(x_n\right)$ and $J\left(T\left(x_n\right)\right)$ converge to the same limit. Hence, by the uniqueness of limits, we have $J T(x)=$ $J(y)$. The injectivity of $J$ plays a crucial role here. Since $J$ is injective, the equation $J T(x)=J(y)$ implies that $T(x)=y$. This demonstrates that $T$ is a closed operator.
By the Closed Graph Theorem, since $T$ is closed, it must be bounded. This completes the proof, showing that the operator $T$ is indeed bounded.
The argument appears to be incomplete where it say "consider a sequence $(x_n)$ in $X$ that converges to $x\in X$. Then $(T(x_n))_n$ converges to some $y \in Y$ and we need to show that $y=T(x)$..." -- the assumption that $(T(x_n))_n$ converges requires justification.
Moreover, I do not think the claim is correct: Suppose that $X$ is a vector space and $\|.\|_1$ and $\|.\|_2$ are norms on $X$ such that $\|v \|_2\leq C\|v\|_1$ for all $v \in X$ and $C$ some positive constant, but that $\{\frac{\|v\|_1}{\|v\|_2}: v \in V\backslash \{0\}\}$ is unbounded in $\mathbb R_{>0}$. If we write $X_1= (X,\|.\|_1)$ and $X_2 = (X,\|.\|_2)$ for the normed vector spaces given by $X$ equipped with $\|.\|_1$ and $\|.\|_2$ respectively, and $\iota_{ij} \colon X_i \to X_j$ ($i,j \in \{1,2\}$) for the identity map on $X$ viewed as a map from $X_i$ to $X_j$, then $\iota_{12},\iota_{11}$ and $\iota_{22}$ are continuous, while $\iota_{21}\colon X_2 \to X_1$ is not. But $\iota_{21}\circ \iota_{12} = \iota_{11}$ is then a counterexample to the claim.
To make the above concrete, one can take, for example, $X = \ell_1 = \{(x_n)_{n \geq 0}: \sum_{n=0}^\infty |x_n| < \infty\}$, the space of absolutely convergent series, with $\|x\|_1 = \sum_{n=0}^\infty |x_n|$, $\|x\|_2 = \sum_{n=0}^\infty |x_n|^2$. Then $\|x\|_2 \leq \|x\|_1$, but if $x^n = (x^n_k)$ where $x_k^n = 1/n$ if $k \leq n$ and $0$ otherwise, then $\|x^n\|_1 = 1$, while $\|x^n\|_2= 1/\sqrt{n}$, and hence $\|x^n\|_1/\|x\|_2 = \sqrt{n}$ so that $\|.\|_1/\|.\|_2$ is not bounded above.