Prove that $T$ is continuous. I am having a really hard time trying to solve this problem:
Let $X$ and $Y$ be normed linear spaces and $T : X \to Y$ a linear operator with closed graph and finite-dimensional range $R(T)$. Prove that T is continuous.
Obviously the closed graph theorem cannot be applied since our spaces are not neccessarily Banach ones. Nonetheless, at least "Y" is a Banach space because of its finite-dimensionality. I do not know how I can go further from here. Thanks in advance. 

Assume $T$ is not continuous at zero. Then there exist a sequence $(x_n)\subset X$ with $x_n\to 0$ and $\epsilon>0$ such that $\|Tx_n\|\ge\epsilon$ for $n\in\mathbb N$. Set $u_n := \frac{x_n}{\|Tx_n\|}\in X$. Then $u_n\to 0$ and $\|Tu_n\|=1$ for all $n\in\mathbb N$. Since $R(T)$ is finite-dimensional, there exists a subsequence $(u_n')$ of $(u_n)$ such that $(Tu_n')$ converges. Let $y\in Y$ be its limit. Then from $u_n'\to 0$ and $Tu_n'\to y$ we conclude that $y=0$. But $\|y\| = \lim_n\|Tu_n'\| = 1$. Contradiction!
