Image of finite-dimensional set is finite-dimensional

Question

Let $X$ be a normed space, $Y,Z\subseteq Z$ linear closed subspaces and let $Y$ be finite-dimensional.

Let $\phi:Y\to X/Z$ be the quotient map. I saw a proof where they say that $\phi(Y)$ is finite-dimensional because $Y$ is.

Is the argument that $\phi$ is bounded, in particular it holds that $||\phi(y)||_{X/Z} \leq ||y||_Y = ||y||_X < \infty$ for every $y\in Y$?

Answer 1

No. It's just an algebraic fact that $\dim F[V]\le \dim V$, because if $\{F(v_i)\}_{i\in I}$ are linearly independent, then $\{v_i\}_{i\in I}$ must be linearly independent too.