Linear application $f$ is discontinuous, then is discontinuous in whole space. 
Let $E,F$ normed vector space. If a linear application $f:E\to F$ is discontinuous, then f is discontinuous in all points of $E$.

My approach: A linear application is a function $f:E\to F$, such that for all $x,y\in E$ and $\lambda\in\mathbb{R}$, we have $f(x+y)=f(x)+f(y)$ and $f(\lambda x)=\lambda f(x)$. So, if f is a discontinuous function, then exist $\epsilon>0$ s.t. for all $\delta>0$, $\vert\vert x\vert\vert<\delta\implies \vert f(x)\vert>\epsilon$, but I don't see how this implies, that f is discontinuous in all points of the vector space E.
 A: Hint: You just have to prove the more general statement:
Let $f\colon G\to G'$ be a group homomorphism, $e$ the neutral element in $G$. Suppose $G$ and $G'$ are endowed with a topology making the group laws continuous. Then $f$ is continuous if and only if it is continuous at $e$.
Vector spaces version:
Let $f\colon E\to F$ be a linear map. Suppose $E$ and $F$ are endowed with a topology making the addition of vectors continuous. Then $f$ is continuous if and only if it is continuous at $0_E$.
A: Proposition : If a linear transform is continuous at any one point then it is continuous every where, this should prove your statement. For this it is sufficient and necessary to prove that if f is continuous at some point then it must be continuous at 0
Proof: Let x be a point in E. let V be any neighborhood of f*0=0 then fx+V is a neighborhood of fx so that , by the continuity of F at x, there is a neighborhood U of x such that f(U) is a subset of fx+v. But f(-x+U) = -fx +f(U) is a subset of V and -x+U is a neighborhood of 0 in E. Hence , f is continuous at 0
A: Let $f \colon E \to F$ be discontinuous at some point, say $x_0$, and assume by contradiction that there is $x_1$ such that $f$ is continuous at $x_1$. Let's write explicitly what this means:

For every open neighborhood $V$ of $f(x_1)$ the is an open neighborhood $U$ of $x_1$ such that $f(U) \subset V$.


Notation: Let $A$ be a subset of the vector space $V$ and let $b$ be a vector in $V$, then we define the translation of the set $A$ by $b$ as $$A + b := \{a + b : a \in A\}.$$ In other words, we add the vector $b$ to each vector of $A$, thus translating the entire set. In the following I am going to use the following fact: let $A$ be an open neighborhood of $b$, then $A - b$ is an open neighborhood of $b - b = 0$.

The plan is to use a translation argument (combined with the linearity of the function $f$) to prove that $f$ is continuous at $x_0$. 
Notice that if we let $W$ be an open neighborhood of $f(x_0)$, $W - f(x_0)$ is an open neighborhood of $0 \in F$ and hence $$W - f(x_0) + f(x_1) = W + f(x_1 - x_0)$$ is an open neighborhood of $f(x_1)$.
Using the property in the yellow box (the continuity at $x_1$), we can find $U$ open such that $x_1 \in U$ and $$f(U) \subset W + f(x_1 - x_0).$$ The right hand side in the previous formula suggests that we translate $U$ by $x_0 - x_1.$ Indeed, $U - x_1 + x_0$ is an open neighborhood of $x_0$, which by construction is mapped inside $W$. To check this we are going to rely heavily on the linearity of $f$, indeed $$f(U - x_1 + x_0) = \color{blue}{f(U) } \color{green}{- f(x_1 - x_0)} \subset \color{blue}{W + f(x_1 - x_0)} \color{green}{- f(x_1 - x_0)} = W.$$
This shows that $f$ is continuous at $x_0$, an obvious contradiction. $\blacksquare$
