Let $G$ be an open subset of $X$ and assume that $f(G)$ is not open. Denoting by $B_m$ the ball with radius $1/m$ around $0$ in $K^n$, this means that there is a $g \in G$ with $ [f(g) + B_m] \cap [K^n\setminus f(G)] \neq \emptyset$ for all $m$, i.e. there is a $c_m = (c_m^{(1)}, ..., c_m^{(n)}) \in B_m$ with $f(g) + c_m \notin f(G)$.
By the surjectivity of $f$, there is some $x_i$ with $f(x_i)=(...,0,1,0,...)$, being $1$ at the $i$-th coordinate. Let now $h_m = g + \sum_{i=1}^n c_m^{(i)} x_i$. Then $f(h_m) = f(g) + c_m \notin f(G)$ for all $m$.
On the other hand, as $(c_m)_m$ converges to $0$ in $K^n$, $(h_m)_m$ converges to $g$ in $X$. As $G$ is an open neighbourhood of $g$, $h_m$ is eventually in $G$ and hence $f(h_m)$ eventually in $f(G)$, contradiction.
To summarize, neither $X$ locally convex nor $f$ continuous is necessary.